Integrals NCERT Solutions
NCERT Exercise 7.5 Q: 16 to 23
Question: 16. `1/(x(x^n+1))`
Solution: `int1/(x(x^n+1))dx`
`=intx^(n-1)/(x^n(x^n+1))dx`
Put `x^n=t`
` =>nx^(n-1)dx=dt`
`=>x^(n-1)dx=1/ndt`
`:. intx^(n-1)/(x^n(x^n+1))dx`
`=1/n int dt/(t(t+1))` ----(i)
The integrand `1/(t(t+1))` is a proper rational function
`:. 1/(t(t+1))=A/t+B/(t+1)` ----(ii)
`=>1=A(t+1)+Bt`
`=> 1=(A+B)t+A`
Equating coefficients of like terms on both sides, we get
`A+B=0` ----(iii)
`A=1` -----(iv)
Putting value of A in (iii), we get
`B=-1`
Putting value of A and B in (ii) we get
`1/(t(t+1))=1/t-1/(t+1)`
`:. int 1/(t(t+1))=int[1/t-1/(t+1)]dt`
`=int 1/tdt-int1/(t+1)dt`
`=log|t|+log|t+1|+C_1`
Putting this value in equation (i), we get
`:. int1/(x(x^n+1))dx`
`=1/n[log|x^n|-log|x^n+1|+C_1]`
`=1/nlog|x^n/(x^n+1)|+1/nC_1`
`=1/nlog|x^n/(x^n+1)|+C`
Where `C=1/nC_1`
Question: 17. `cosx/((1-sinx)(2-sinx))`
Solution: `int cosx/((1-sinx)(2-sinx))dx`
Put `sinx=t =>cosxdx=dt`
`:.int cosx/((1-sinx)(2-sinx))dx`
`=int dt/((1-t)(2-t))`
The integrand `1/((1-t)(2-t))` is a proper rational function
`:. 1/((1-t)(2-t))=A/(1-t)+B/(2-t)` ----(i)
`=>1=A(2-t)+B(1-t)`
`=> 1=(-A-B)t+(2A+B)`
Equating coefficients of like terms on both sides
`-A-B=0`
`=>A+B=0` ----- (ii)
`2A+B=1` ----(iii)
Subtracting (ii) from (iii), we get
`A=1`
Putting values of A in (ii), we get
`B=-1`
Putting values of A and B in (i), we get
`1/((1-t)(2-t))=1/(1-t)+1/(2-t)`
`:.int dt/((1-t)(2-t))` `=int[1/(1-t)+1/(2-t)]dt`
`=int1/(1-t)dt-int1/(2-t)dt`
`=log|1-t|/(-1)-log|2-t|/(-1)+C`
`=-log|1-sinx|+log|2-sinx|+C`
`=log|(2-sinx)/(1-sinx)|+C`
Question: 18. `((x^2+1)(x^2+2))/((x^2+3)(x^2+4))`
Solution: Let `I=int((x^2+1)(x^2+2))/((x^2+3)(x^2+4))dx`
Consider `((x^2+1)(x^2+2))/((x^2+3)(x^2+4))`
Let, `x^2=t=((t+1)(t+2))/((t+3)(t+4)`
`=(t^2+3t+2)/(t^2+7t+12)=1+(-4t-10)/(t^2+7t+12)`
`=1-[((4t+10))/((t+3)(t+4))]`
Let `(4t+10)/((t+3)(t+4))=A/(t+3)+B/(t+4)`
`=>4t+10=A(t+4)+B(t+3)`
Let, `t+3=0=>t=-3`
`:.4(-3)+10=A(-3+4)`
`=>A=-2`
Again let `t+4=0=>t=-4`
`:.4(-4)+10=B(-4+3)=>B=6`
`:.((x^2+1)(x^2+2))/((x^2+3)(x^2+4))` `=1-[(-2)/(t+3)+6/(t+4)]`
`=1+[2/(x^2+3)-6/(x^2+4)]`
`:. I=int[1+2/(x^2+3)-6/(x^2+4)]`
`=x+2/sqrt3 tan^-1` `x/sqrt3-6/2tan^-1` `x/2+C`
`=x+2/sqrt3 tan^-1` `x/sqrt3-3tan^-1` `x/2+C`
Question: 19. `(2x)/((x^2+1)(x^2+3))`
Solution: Let `I=int (2x)/((x^2+1)(x^2+3))dx`
Put `x^2=y`, so that `2xdx=dy`
`:. I=int dy/((y+1)(y+3))`
Let `1/((y+1)(y+3))=A/(y+1)+B/(y+3)`
`=>1=A(y+3)+B(y+1)` ---(i)
Putting `y=-1` in (i), we get
`1=2A=>A=1/2`
Putting `y=-3` in (i), we get
`1=-2B=>B=-1/2`
`:.1/((y+1)(y+3))=1/(2(y+1))-1/(2(y+3))`
`:. I=int [1/(2(y+1))-1/(2(y+3))]dy`
`=1/2intdy/(y+1)-1/2intdy/(y+3)`
`=1/2log|y+1|-1/2log|y+3|+C`
`=1/2log|(y+1)/(y+3)|+C`
`=1/2|(x^2+1)/(x^2+3)|+C`
Question: 20. `1/(x(x^4-1))`
Solution: Let, `I =int1/(x(x^4-1))dx`
`=1/4int(4x^3dx)/(x^4(x^4-1))`
Put `x^4=t`, so that `4x^3dx=dt`
`:.I=1/4intdt/(t(t-1))`
Let, `1/(t(t-1))-=A/t+B/(t-1)`
`=>1-=A(t-1)+Bt` ---(i)
Putting `t=0` in (i), we get
`A=-1`
Putting `t=1` in equation (i), we get
`B=1`
`:. 1/(t(t-1))=-1/t+1/(t-1)`
`:.I=1/4int(-1/t+1/(t-1))dt`
`=1/4[-log|t|+log|t-1|+C`
`=1/4log|(t-1)/t|+C`
`=1/4log|(x^4-1)/x^4|+C`
Question: 21. `1/(e^x-1)`
Solution: Let, `I=1/(e^x-1)dx`
Let, `e^x=t =>e^xdx=dt`
`=>dx=(dt)/t`
`:. I=int(dt)/(t(t-1))`
Let, `1/(t(t-1))=A/t+B/(t-1)`
`=>I=A(t-1)+Bt`
Let, `t=0=>1=-A`
`=>A=-1`
Let, `t-1=0=>t=1`
`=>B=1`
`:. I=int((-1)/t+1/(t-1))dt`
`=-logt+log(t-1)+C`
`=-loge^x+log(e^x-1)+C`
`=log((e^x-1)/e^x)+C`
Choose the correct answer in each of the following:
Question: 22. `int (xdx)/((x-1)(x-2))` equals
(A) `log|(x-1)^2/(x-2)|+C`
(B) `log|(x-2)^2/(x-1)|+C`
(C) `log|(x-1^2)/(x-2)|+C`
(D) `log|(x-1)(x-2)|+C`
Answer: Option (B) `log|(x-2)^2/(x-1)|+C`
Explanation
Given, `int (xdx)/((x-1)(x-2))`
`=int[(-1)/(x-1)+2/(x-2)]dx`
`=-log(x-1)+2log(x-2)+C`
`=log|(x-2)^2/(x-1)|+C`
Thus, Option (B) `log|(x-2)^2/(x-1)|+C` is correct answer
Question: 23. `int(dx)/(x(x^2+1))` equals
(A) `log|x|-1/2log(x^2+1)+C`
(B) `log|x|+1/2log(x^2+1)+C`
(C) `-log|x|+1/2log(x^2+1)+C`
(D) `1/2log|x|+log(x^2+1)+C`
Answer: option(A) `log|x|-1/2log(x^2+1)+C`
Explanation:
Let, `1/(x(x^2+1))=A/x+(Bx+Cx)/(x^2+1)`
`=>1=A(x^2+1)+(Bx+C)(x)`
Let,`x=0`
`:. 1=A=>A=1`
Comapring coefficients of `x^2`
`0=A+B=>B=-1`
Comparing coefficients of x
`0=C=>C=0`
`:.int1/(x(x^2+1))dx` `=int[1/x+(-x)/(x^2+1)]dx`
`=logx-1/2log(x^2+1)+C`
Thus, option(A) `log|x|-1/2log(x^2+1)+C` is the correct answer.
Reference: