NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.3) Exercise 7.3

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NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.3) Exercise 7.3 

The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.3 deals with the exercise on the topic of Definite and Indefinite Integrals and their properties. Integrals include two major concepts – the problem of finding the function whenever the derivatives are given, and the problem of finding the area bounded by the graph function under certain conditions. The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.3 contain Integration which is called the inverse process of Differentiation. Instead of differentiating a function, the derivative function is given, and the question asks for its primitive. For a better understanding of Integrals, one should refer to the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.3. Practising the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.3 allows students to build a strong conceptual foundation for a better understanding of mathematical solutions. It covers all important theorems and formulae with detailed explanations.

The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.3 explain Integration using some trigonometric functions, and a few specific identities are used to find the integral. The NCERT Solutions for Exercise 7.3 Class 12th are focused on covering various determinants of Integrals. Class 12 Maths Ex 7.3 is said to be a significant exercise, as the same pattern of questions can be observed in past years’ board examinations. Hence, students should refer to the NCERT Solutions For Class12 Maths Exercise 7.3 for improving their scores. Consistency is the key to success, so for scoring well in examinations, students must revise the NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.3 consistently. Extramarks provides detailed NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.3 that help in understanding the concepts, thereby helping in improving grades.

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CBSE Class 12 Maths NCERT Solutions For Chapter 7 Exercise 7.3

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The NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.3 cover the topic of Integration while also using some trigonometric functions. Trigonometric functions are not just used while solving NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7, they can be used in a variety of topics. Therefore, mathematics is not just a subject that a student needs to learn formulas for, aiming for a good score, but it is also necessary for the future and daily planning of any individual.

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Evaluate the following definite integrals as limit of sums.23x2dx


We have, 23x2dxabf(x)dx=(ba)limn1n[f(a)+f(a+h)+...+f(a+(n1)h)],where,   h=banHere,  a=2,  b=3,  f(x)=x2,  h=32n=1n23x2dx          =(32)limn1n[f(2)+f(2+h)+...+f(2+(n1)h)]          =limn1n[(2)2+(2+h)2+...+{2+(n1)h}2]          =limn1n[4+(4+4h+h2)+...+{4+4(n1)h+(n1)2h2}]          =limn1n[4n+4{1+2+...+(n1)}h+{1+22+32+...+(n1)2}h2]          =limn1n[4n+4(n1)n2h+(n1)n(2n2+1)6h2]          =limn1n[4n+4(n1)n2.1n+(n1)n(2n2+1)6.1n2]                                        [n=n(n+1)2,n2=n(n+1)(2n+1)6]          =limn[4+4(11n)2+(11n)(21n)6]          =[4+4(10)2+(10)(20)6]          =4+2+13          =193


Evaluate the following definite integrals as limit of sums.05(x+1)dx


Wehave,  05(x+1)dxBy definitionabf(x)dx=(ba)limn1n[f(a)+f(a+h)+...+f(a+(n1)h)],where,   h=banHere,  a=0,  b=5,  f(x)=x+1,  h=50n=5n05(x+1)dx=(ba)limn1n[f(0)+f(0+h)+...+f(0+(n1)h)]                            =5limn1n[(0+1)+(h+1)+...+{(n1)h+1}]                            =5limn1n[n+h+2h+...+(n1)h]                            =5limn1n[n+{1+2+...+(n1)}h]                            =5limn1n[n+(n1)n2h][n=n(n+1)2]                            =5limn[1+(n1)2(5n)]                            =5limn[1+52(11n)]                            =5[1+52(10)]                            =5×72  =352


Evaluate the following definite integrals as limit of sums.abxdx


By definitionabf(x)dx=(ba)limn1n[f(a)+f(a+h)+...+f(a+(n1)h)],where,   h=banHere,  f(x)=x,  h=banabxdx=(ba)limn1n[f(a)+f(a+h)+...+f(a+(n1)h)]          =(ba)limn1n[a+a+h+...+a+(n1)h]          =(ba)limn1n[na+h+2h+...+(n1)h]          =(ba)limn1n[na+{1+2+...+(n1)}h]          =(ba)limn1n[na+(n1)n2h]        [n=n(n+1)2]          =(ba)limn[a+(n1)2(ban)]          =(ba)limn[a+(ba)2(11n)]          =(ba)[a+(ba)2(10)]          =(ba)(2a+ba2)          =(ba)(b+a)2  =b2a22




(x+3)34xx2dxLet x+3=Addx(34xx2)+B        x+3=A(42x)+B          3=4A+B   & 2A=1    A=12 &     B=3+4A       =3+4(12)       =32=1(x+3)34xx2dx=12(42x)34xx2dx                                          +134xx2dx                              =12(34xx2)3232+7(x+2)2dx                              =(34xx2)323+(7)2(x+2)2dx                         =(34xx2)323+12(x+2)(7)2(x+2)2+72sin1(x+27)+C                                                              +(7)22sin1(x+27)+C                    =(34xx2)323+12(x+2)34xx2  +72sin1(x+27)+C




(x+1)2x2+3dxLet x+1=Addx(2x2+3)+B        x+1=A(4x+0)+B    x+1=4Ax   & B=1    A=14 & B=1(x+1)2x2+3dx=144x2x2+3dx+12x2+3dx                       =14×(2x2+3)3232+2x2+(32)2dx                =16×(2x2+3)32+2x2+(32)2dx                =16×(2x2+3)32+2(12xx2+(32)2        +  (32)22log|x+(32)2|)+C                =16×(2x2+3)32+22xx2+32                                    +324log|x+x2+(32)2|+C                =(2x2+3)326x22x2+3324log|x+x2+32|+C




xx+x2dxLet x=Addx(x+x2)+B       x   =A(1+2x)+B     1=2A & A+B=0    A=12 & B=12xx+x2dx=12(1+x)x+x2dx12x+x2dx               =12×(x+x2)32(32)12(x+12)2(12)2dx               =13×(x+x2)3212×(x+12)2(x+12)2(12)2                              +12×(12)22log|x+(x+12)2(12)2|+C               =(x+x2)32318(2x+1)x+x2+116log|x+x+x2|+C


Choosethecorrectanswerx28x+7dx is equal toA12x4x28x+7+9logx4+x28x+7+CB14x+4x28x+7+9logx+4+x28x+7+CC12x4x28x+732logx4+x28x+7+CD12x4x28x+792logx4+x28x+7+C


We have,  x28x+7=(x4)29Let I=(x4)2(3)2dx          =t2(3)2dt     [Let t=x4dtdx=1]                            [x2a2dx=x2x2a2a22log|x+x2a2|]        =t2t2(3)2(3)22log|t+t2(3)2|+CPutting t=x4,weget     =(x4)2(x4)2(3)292log|(x4)+(x4)2(3)2|+C     =(x4)2x28x+792log|(x4)+x28x+7|+CHence, the correct option is D.


Choosethecorrectanswer1+x2dx is equal toAx21+x2+12logx+1+x2+CB231+x232+CC23x1+x232+CDx221+x2+12 x2logx+1+x2+C


Let  I=1+x2dx          =x2+12dx                        [x2+a2dx=x2x2+a2+a22log|x+x2+a2|]           =(x2x2+12+122log|x+x2+12|)+C           =x2x2+1+12log|x+x2+1|+C           =x21+x2+12log|x+1+x2|+CHence,​  the correct option is A.




Wehave,1+x29=139+x2Let  139+x2dx=1332+x2dx                                           =13x2+32dx                                                        [x2+a2dx=x2x2+a2+a22log|x+x2+a2|]                       =13(x2x2+32+322log|x+x2+32|)+C                       =x6x2+9+32log|x+x2+9|+C




We have,  x2+3x=(x2+3x+94)94                                          =(x+32)2(32)2Let I=(x+32)2(32)2dx         =t2(32)2dt     [Let t=x+32dtdx=1]                                 [x2a2dx=x2x2a2a22log|x+x2a2|]         =t2t2(32)2(32)22log|t+t2(32)2|+CPutting t=x+32, we get     =(x+32)2(x+32)2(32)298log|(x+32)+(x+32)2(32)2|+C     =(2x+3)4x2+3x98log|(x+32)+x2+3x|+C




We have,  1+3xx2=1(x23x+9494)                                                   =(132)2(x32)2Let I=(132)2(x32)2dx         =(132)2t2dt      [Let t=x32dtdx=1]                  [a2x2dx=x2a2x2+a22sin1xa]      =t2(132)2t2+(132)22sin1t(132)+CPutting t=x32, we get      =(x32)2(132)2(x32)2+(132)22sin1(x32)(132)+C      =(2x3)41+3xx2+138sin1(2x3)(13)+C




We have,  x2+4x5=(x+2)29Let I=(x+2)2(3)2dx         =t2(3)2dt      [Let t=x+2dtdx=1]                                 [x2a2dx=x2x2a2a22log|x+x2a2|]      =t2t2(3)2(3)22log|t+t2(3)2|+CPutting t=x+2, we get      =(x+2)2(x+2)2(3)292log|(x+2)+(x+2)2(3)2|+C      =(x+2)2x2+4x592log|(x+2)+x2+4x5|+C




We have,  14xx2=5(x2+4x+4)                                                    =(5)2(x+2)2Let I=(5)2(x+2)2dx         =(5)2t2dt     [Let t=x+2dtdx=1]                      [a2x2dx=x2a2x2+a22sin1xa]      =t2(5)2t2+(5)22sin1t(5)+CPutting t=x+2,weget      =(x+2)2(5)2(x+2)2+(5)22sin1(x+2)(5)+C      =(x+2)214xx2+52sin1(x+2)(5)+C




We have,  x2+4x+1=(x+2)23Let I=(x+2)2(3)2dx         =t2(3)2dt     [Let t=x+2dtdx=1]                      [x2a2dx=x2x2a2a22log|x+x2a2|]       =t2t2(3)2(3)22log|t+t2(3)2|+CPutting t=x+2, we get       =(x+2)2(x+2)2(3)232log|(x+2)+(x+2)2(3)2|+C       =(x+2)2x2+4x+132log|(x+2)+x2+4x+1|+C




We have,  x2+4x+6=(x+2)2+2Let I=(x+2)2+(2)2dx         =t2+(2)2dt   [Let t=x+2dtdx=1]                                      [x2+a2dx=x2x2+a2+a22log|x+x2+a2|]       =t2t2+(2)2+(2)22log|t+t2+(2)2|+CPutting t=x+2, we get     =(x+2)2(x+2)2+(2)2+22log|(x+2)+(x+2)2+(2)2|+C     =(x+2)2x2+4x+6+log|(x+2)+x2+4x+6|+</mo&g




We have,  14x2Let I=14x2dx          =12(2x)2dx[Let t=2xdtdx=2dx=dt2]            =12t2dt2                             [a2x2dx=12xa2x2+a22sin1xa+C]           =12(12t12t2+a22sin1t1)+C           =12(2x214x2+12sin12x1)+C          =x214x2+14sin12x+C




We have,  4x2Let I=4x2dx           =22x2dx                         [a2x2dx=12xa2x2+a22sin1xa+C]           =12x22x2+222sin1x2+C          =x24x2+2sin1x2+C


Choosethecorrectanswerexsecx1+tanxdxequalsA​ excosx+CBexsecx+CCexsinx+CDextanx+C


Let  I=exsecx(1+tanx)dx          =ex(secx+secxtanx)dxLet​ f(x)=secx  and f’(x)=secxtanxThus, the given integrand is of the form ex[f(x)+f(x)].Therefore,       I=ex(secx+secxtanx)dx        =exsecx+C       [ex{f(x)+f(x)}=exf(x)]Hence, the correct option is B.


Choosethecorrectanswerx2ex3dxequalsA​ 13ex3+CB13ex2+CC12ex3+CD12ex2+C


We  have  x2ex3dxLet​ t=x3dtdx=3x2x2ex3dx=x2etdt3x2                          =13etdt                          =13ex3+CHence, the correct option is A.


Integrate the functionssin1(2x1+x2)


We have, sin1(2x1+x2)Let  I=sin1(2x1+x2)dxLetx=tanθdx=sec2θ     I=sin1(2x1+x2)dx          =sin1(2tanθ1+tan2θ)sec2θ          =sin1(sin2θ)sec2θ          =2θsec2θ         =2[θsec2θ(dθsec2θ)]         =2[θtanθ1tanθ]         =2θtanθ2log|secθ|+C          =2θtanθ2log|1+tan2θ|+C       I=2xtan1x2log|1+x2|+C.

Q21. Integrate the functions e2x sinx


We have, e2xsinxLet  I=e2xsinxdxTaking sinx as first function and e2x as second function, then Integrating by parts, we get      I=e2xsinxdx        =sinxe2xdx(ddxsinxe2xdx)dx        =sinx(e2x2){cosx(e2x2)}dx        =e2x2sinx12cosx.e2xdx        =e2x2sinx12{cosxe2xdx(ddxcosxe2xdx)dx}        =e2x2sinx12{cosx(e2x2){sinx(e2x2)}dx}        =e2x2sinx12cosx(e2x2)14e2xsinxdx+C     I=e2xsinx214cosxe2x14I+CI+14I=e2xsinx214cosxe2x+C      54I=e2xsinx214cosxe2x+C         I=45e2x4(2sinxcosx)+C


Integrate the functions(x3)ex(x1)3


Let  I=(x3)ex(x1)3dx           =ex{x12(1x)3}dx           =ex{x1(1x)32(1x)3}dx           =ex{1(1x)22(1x)3}dxLet​ f(x)=1(1x)2  and f’(x)=2(1x)3Thus, the given integrand is of the form ex[f(x)+f(x)].Therefore,       I=(x3)ex(x1)3dx        =ex{1(1x)22(1x)3}dx       =ex(1x)2+C            [ex{f(x)+f(x)}=exf(x)]


Integrate the functionse x(1x1x2)


Let  I=ex(1x1x2)dxLet​ f(x)=1x  and f’(x)=1x2Thus, the given integrand is of the form ex[f(x)+f(x)].Therefore,   I=ex(1x1x2)dx                            =exx+C           [ex{f(x)+f(x)}=exf(x)]


Integrate the functions ex(1+sinx1+cosx)


We have, ex(1+sinx1+cosx)=ex(sin2x2+cos2x2+2sinx2cosx22cos2x2)                           =12ex(sinx2+cosx2cosx2)2                           =12ex(tanx2+1)2                           =12ex(1+tan2x2+2tanx2)                           =12ex(sec2x2+2tanx2)                          =ex(12sec2x2+tanx2)                          =ex(tanx2+12sec2x2)Let f(x)=tanx2  and f’(x)=12sec2x2Thus, the given integrand is of the form ex[f(x)+f(x)].Therefore,ex(1+sinx1+cosx)dx=ex(tanx2+12sec2x2)dx                                           =extanx2+C    [ex{f(x)+f(x)}=exf(x)]


Integrate the functions xex(1+x)2


Let  I=xex(1+x)2dx           =ex{x(1+x)2}dx          =ex{1+x1(1+x)2}dx         =ex{1+x(1+x)21(1+x)2}dx          =ex{1(1+x)1(1+x)2}dxLet​ f(x)=1(1+x)  and f’(x)=1(1+x)2Thus, the given integrand is of the form ex[f(x)+f(x)].Therefore,       I=xex(1+x)2dx        =ex{1(1+x)1(1+x)2}dx       =ex1+x+C            [ex{f(x)+f(x)}=exf(x)]

Q26. Integrate the functions ex (sinx + cosx)


Let  I=ex(sinx+cosx)dxHere, f(x)=sinx and f’(x)=cosxThus, the given integrand is of the form ex[f(x)+f(x)].Therefore,       I=ex(sinx+cosx)dx        =exsinx+C       [ex{f(x)+f(x)}dx=exf(x)+C]

Q27. Integrate the functions (x2+1) logx


Let  I=(x2+1)logxdxTaking logx as a first function and (x2+1) as second function and integrating by parts, we get      =logx(x2+1)dx{(ddxlogx)(x2+1)dx}dx      =logx(x33+x)1x.(x33+x)dx      =(x33+x)logx(x23+1)dx+C      =(x33+x)logxx39x+C

Q28. Integrate the functions x (logx)2


Let  I=x(logx)2dxTaking (logx)2 as a first function and x as second function and integrating by parts, we get      =(logx)2xdx{(ddx(logx)2)xdx}dx      =(logx)2(x22)2logxx.(x22)dx      =x22(logx)2xlogxdx+C      =x22(logx)2[logxxdx{(ddxlogx)xdx}dx]      =x22(logx)2{logx(x22)1x.(x22)dx}      =x22(logx)2{x22logxx2dx}      =x22(logx)2x22logx+x24+C

Q29. Integrate the functions tan-1x


Let  I=tan1xdx=tan1x.1dxTaking tan1x as a first function and 1 as second function and integrating by parts, we get         =tan1x1dx{(ddxtan1x)1dx}dx         =tan1x(x)11+x2.(x)dx         =xtan1xx1+x2dx+C         =xtan1x12log|1+x2|+C 

Q30. Integrate the functions x sec2x


Let  I=xsec2xdxTaking algebraic function i.e. x as a first function and sec2x assecond function and integrating by parts, we get         =xsec2xdx{(ddxx)sec2xdx}dx        =x(tanx)1.tanxdx         =xtanxtanxdx+C         =xtanx+log|cosx|+C


Integrate the functionsxcos1x1x2


Let  I=xcos1x1x2dx          =12cos1x2x1x2dxTakingcos1x as first function and 2x1x2 as second function and integrating by parts, we get        I=12{cos1x2x1x2dx(ddxcos1x2x1x2dx)dx}         =12{cos1x2xtdt2x(ddxcos1x2xtdt2x)dx}                                              [Let t=1x2dtdx=2xdx=dt2x]        =12[cos1x.2t(11x22t)dx]        =12[cos1x.21x2(11x221x2)dx]        =12(2cos1x1x2+21dx)       =12(2cos1x1x2+2x)+C       =(1x2cos1x+x)+C

Q32. Integrate the functions


Let  I=(sin1x)2dx          =(sin1x)2.1dxTaking(sin1x)2 as first function and 1 as second function and integrating by parts, we get        I=(sin1x)21dx{ddx(sin1x)21dx}dx      =(sin1x)2x{2sin1xddx(sin1x).x}dx      =(sin1x)2x2xsin1x1x2dx      =(sin1x)2x+sin1x2x1x2dx      =(sin1x)2x+sin1x2x1x2dx(ddxsin1x2x1x2dx)dx      =(sin1x)2x+sin1x2xtdt2x{ddxsin1x2xtdt2x}dx                                                        [Let t=1x2dtdx=2xdx=dt2x]      =(sin1x)2x+sin1x.2t{11x22t}dx      =(sin1x)2x+2sin1x.1x211x221x2dx      =(sin1x)2x+2sin1x.1x221dx      =x(sin1x)2+21x2sin1x2x+C

Q33. Integrate the functions
x cos-1x


Let  I=xcos1xdxLet t=cos1xx=cost  sint=1cos2t=1x2Then,  dtdx=ddxcos1x                       =11x2                      =1sint         dx=sintdtSo,             I=tcost(sint)dt                      =12tsin2tdtTaking algebraic function i.e. t as a first function and sin2t as second function and integrating by parts, we get     =12[tsin2tdt{(ddtt)sin2tdt}dt]     =12[t(12cos2t)1.(12co2t)dx]     =12[12tcos2t+14sin2t+C]     =14t(12sin2t)18×2sintcost+C     =14t(12sin2t)14sintcost+C     =14(12x2)sin1x14x1x2+C

Q34. Integrate the functions
x tan-1x


Let  I=xtan1xdxTaking tan1x as a first function and x as second function and integrating by parts, we get=tan1xxdx{(ddxtan1x)xdx}dx=tan1x(x22)11+x2.(x22)dx=x22tan1x12x21+x2dx+C=x22tan1x121+x211+x2dx+C=x22tan1x121+x21+x2dx+1211+x2dx+C=x22tan1x121.dx+12tan1x+C=x22tan1x12xdx+12tan1x+C

Q35. Integrate the functions
x sin-1x


Let  I=xsin1xdxLet t=sin1xx=sintcost=1sin2t=1x2Then,  dtdx=ddxsin1x                    =11x2                   =1cost        dx=costdtSo,          I=tsintcostdt               =12tsin2tdtTaking algebraic function i.e. t as a first function and sin2t as second function and integrating by parts, we get      =12[tsin2tdt{(ddtt)sin2tdt}dt]      =12[t(12cos2t)1.(12co2t)dx]      =12[12tcos2t+14sin2t+C]      =14t(12sin2t)+18×2sintcost+C      =14t(2sin2t1)+14sintcost+C      =14(2x21)sin1x+14x1x2+C

Q36. Integrate the functions
x2 logx


Let  I=x2logxdxTaking logx as a first function and x2 as second function and integrating by parts, we get     =logxx2dx{(ddxlogx)x2dx}dx      =logx(x33)1x.(x33)dx      =x33logxx23dx+C      =x33logxx39+C

Q37. Integrate the functions
x log 2x


Let  I=xlog2xdxTaking log2x as a first function and x as second function and integrating by parts, we get     =log2xxdx{(ddxlog2x)xdx}dx     =log2x(x22)12x×2.(x22)dx     =x22log2xx2dx+C     =x22log2xx24+C

Q38. Integrate the functions
x logx


Let  I=xlogxdxTaking logx as a first function and x as second function and integrating by parts, we get =logxxdx{(ddxlogx)xdx}dx =logx(x22)1x.(x22)dx   =x22logxx2dx+C      =x22logxx24+C

Q39. Integrate the functions
x2 ex


Let  I=x2ex< /msup>dxTaking algebraic function i.e. x2 as a first function and ex as second function and integrating by parts, we get     =x2exdx{(ddxx2)exdx}dx     =x2(ex)2x.(ex)dx     =x2ex2[xexdx{(ddxx)exdx}dx]+C     =x2ex2[xex1.exdx]+C     =x2ex2xex+2ex+C     =ex(x22x+2)+C

Q40. Integrate the functions
x sin 3x


Let  I=xsin3xdxTaking algebraic function i.e. x as a first function and sin3x as second function and integrating by parts, we get      =xsin3xdx{(ddxx)sin3xdx}dx      =x(13cos3x)1.(13cosx)dx      =13xcos3x+19sin3x+C

Q41. Integrate the function x sin x.


Let  I=xsinxdxTaking algebraic function i.e. x as a first function and sinx as second function and integrating by parts, we get=xsinxdx{(ddxx)sinxdx}dx=x(cosx)1.(cosx)dx=xcosx+sinx+C


Choosethecorrectanswerdxxx2+1equalsA​ logx12logx2+1+CBlogx+12logx2+1+CClogx+12logx2+1+CD12logx+logx2+1+C


We have,  dxx(x2+1)Let 1x(x2+1)=Ax+Bx+C(x2+1)                          =A(x2+1)+(Bx+C)xx(x2+1)               1=A(x2+1)+(Bx2+Cx)               1=x2(A+B)+Cx+AEquating the coefficients of x2, x and constant term, we getA+B=0,   C=0 and A=1On solving these equations, we getA=1,  B=1 and C=0             1x(x2+1)=1x+x(x2+1)1x(x2+1)dx=1xdxx(x2+1)dx                                       = log|x|2log|x2+1|+CThus, the correct option is (A).


Choosethecorrectanswerxdxx1x2dxequalsA​ logx12x2+CBlogx22x1+CClogx1x22+CDlogx1x2+C


We have,  xdx(x1)(x2)dxLet x(x1)(x2)=Ax1+Bx2                                    =A(x2)+B(x1)(x1)(x2)                         x=A(x2)+B(x1)    ...(i)Substituting x=1  and  2, in equation(i),​ we getA=1 and B=2                x(x1)(x2)=1x1+2x2x(x1)(x2)dx=1x1dx+2x2dx                                                 = log|x1|+2log|x2|+C                                                =log|(x2)2(x1)|+CHence, the option (B) is correct.


Integrate the rational function 1(ex1)


1(ex1)dxLet  t=exdtdx=ex     dx=dtex=dtt1(ex1)dx=1(t1)dtt                                  =1t(t1)dtLet  1t(t1)=At+Bt1                       =A(t1)+Btt(t1)              1=A(t1)+Bt     ...(i)Putting t=0​ and 1 respectively in equation (i),we getA=1 and B=1             1t(t1)=1t+1t11t(t1)dt=1tdt+1t1dt                                    =log|t|+log|t1|+C                                    =log|t1t|+C1(ex1)dx=log|ex1ex|+C


Integrate the rational function 1x(x41)


We have, 1x(x41)=1x(x41)×x3x3                                         =x3x4(x41)Let  t=x4dtdx=4x41dt4x3=dx1x(x41)dx=x3x4(x41)dx                                     =x3t(t1)dt4x3                                     =141t(t1)dtLet 1t(t1)=At+Bt1                     1=A(t1)+BtSubstituting​ t= 0 and 1 respectively, we getA=1  and   B=1∴                 1t(t1)=1t+1t11x(x41)dx=14(1t+1t1)dt                                       =141tdt+141t1dt                                         =14log|t|+14log|t1|+C                                         =14log|x4|+14log|x41|+C                                         =14log|x41x4|+C


Integrate the rational function 2x(x2+1)(x2+3)


We have,  2x(x2+1)(x2+3)Let  t=x2dtdx=2xdx=dt2x2x(x2+1)(x2+3)dx=2x(t+1)(t+3)dt2x                                               =1(t+1)(t+3)dtLet  1(t+1)(t+3)=At+1+Bt+3                               1=A(t+3)+B(t+1)Substituting t=1 and3 respectivelywe getA=12  and  B=121(t+1)(t+3)=12(t+1)12(t+3)1(t+1)(t+3)dx=12(t+1)dt12(t+3)dx                                             =12log|t+1|12log|t+3|+C                                             =12log|t+1t+3|+C                                               =12log|x2+1x2+3|+C


Integrate the rational function (x2+1)(x2+2)(x2+3)(x2+4)


We have,   (x2+1)(x2+2)(x2+3)(x2+4)=1+(4x2+10)(x2+3)(x2+4)Let,(4x2+10)(x2+3)(x2+4)=Ax+B(x2+3)+Cx+D(x2+4)             4x2+10=(Ax+B)(x2+4)+(Cx+D)(x2+3)                                =x3(A+C)+x2(B+D)+x(4A+3C)+(4B+3D)Comparing the coefficients of x3, x2, x and constant terms,we get   A+C=0B+D=44A+3C=04B+3D=10On​ solving these equations, we getA=0,B=2,C=0 and D=6(4x2+10)(x2+3)(x2+4)=2(x2+3)+6(x2+4)Then,(x2+1)(x2+2)(x2+3)(x2+4)dx=1dx(2(x2+3)dx+61(x2+4)dx)                                                 =x+23tan1(x3)62tan1(x2)+C                                                =x+23tan1(x3)3tan1(x2)+C


Integrate the rational function cosx(1sinx)(2sinx)


We have, cosx(1sinx)(2sinx)Let  t=sinxdtdx=cosxdtcosx=dxcosx(1sinx)(2sinx)dx=cosx(1t)(2t)dtcosx                                                               =1(1t)(2t)dtLet  1(1t)(2t)=A1t+B2t                                1=A(2t)+B(1t)     ...(1)Substituting t=1 and 2 respectively in eqution (1),we getA=1 and B=11(1t)(2t)=11t12tSo,cosx(1sinx)(2sinx)dx=1(1t)(2t)dt                                                 =11tdt12tdt                                                  =log|1t|+log|2t|+C                                                 =log|2t1t|+C                                                  =log|2sinx1sinx|+C


Integrate the rational function 1x(xn1)


We have, 1x(xn1)=1x(xn1)×xn1xn1                                          =xn1xn(xn1)Let    t=xndtdx=nxn1dtnxn1=dx1x(xn1)dx=xn1xn(xn1)dx                                       =1n1t(t1)dtLet 1t(t1)=At+Bt1                       1=A(t1)+Bt                      =t(A+B)ASubstituting​ t= 0 and 1 respectively, we getA=1  and   B=1          1t(t1)=1t+1t11x(xn1)dx=1n(1t1t1)dt                                      =1n1tdt1n1t1dt                                     =1nlog|t|1nlog|t1|+C                                     =1nlog|xn|1nlog|xn1|+C                                    =1nlog|xnxn1|+C


Integrate the rational function 1x41


We have, 1x41=1(x+1)(x1)(x2+1)Let  1x41=A(x+1)+B(x1)+Cx+D(x2+1)                   1=A(x1)(x2+1)+B(x+1)(x2+1)+(x21)(Cx+D)                   1=A(x3+xx21)+B(x3+x+x2+1)+C(x3x)+D(x21)                   1=x3(A+B+C)+x2(A+B+D)+x(A+BC)+(A+BD)Equating the coefficients of x2, x and constant term, we get       A+B+C=0A+B+D=0       A+BC=0A+BD=1Solving, the above equations, we getA=14,  B=14,C=0 and D=12    1x41=14(x+1)+14(x1)+12(x2+1)1x41dx=141x+1dx+141x1dx121x2+1dx                                 =14log|(x+1)|+14log|(x1)|12tan1x+C                                 =14log|x1x+1|12tan1x+C

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