Math Twelve

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.

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