Math Twelve

Integrals NCERT Solutions

NCERT Exercise 7.5 Q: 8 to 15

Question:8.`x/((x-1)^2(x+2))dx`

Solution:

`intx/((x-1)^2(x+2))dx`

The integrant` x/((x-1)^2(x+2))dx` is a proper rational function.

`:.x/((x-1)^2(x+2))` `=A/(x-1)+ B/((x-1)^2)+C/(x+2)`

`=>x=A(x-1)(x+2)` `+B(x+2)+C(x-1)^2`

`=>x=A(x^2+x-2)` `+B(x+2)+C(x-1)^2`

`=>x=(A+C)x^2+(A+B-2C)x` `+(-2A+2B+C)`

Equating coefficients of like terms on both sides, we have

`A + C = 0` ----(ii)

`A+B-2C=1` -----(iii)

`-2A+2B+C=0` ----(iv)

Multiplying (ii)by 2 and then subtracting from (iv), we get

`-4A+5C=-2` ------(v)

Multiplying (ii) by 4 and adding with (v), we get

`C=-2/9`

Putting value of C in (ii), we get

`A=2/9`

Putting values of C in (iii) we get

`B=1/3`

Putting the values of A, B and C in (i), we get

`x/((x-1)^2(x+2))` `=2/(9(x-1))+1/(3(x-1)^2)-2/(9(x+2)`

`:. intx/((x-1)^2(x+2))dx`

`=int [2/(9(x-1))+1/(3(x-1)^2)-2/(9(x+2))]dx`

`=2/9int1/((x-1))dx+1/3int1/((x-1)^2)dx` `-2/9int1/(x+2)dx`

`=2/9log|x-1|+1/3((x-1)^-1)/(-1)` `-2/9log|x+2|+C`

`=2/9log|x-1|-1/(3(x-1))` `-2/9log|x+2|+C`

`=2/9log|(x-1)/(x+2)|-1/(3(x-1))+C`

Question: 9. `(3x+5)/(x^3-x^2-x+1)`

Solution:

`int(3x+5)/(x^3-x^2-x+1)dx`

`=int(3x+5)/((x-1)^2(x+1))dx`

The integrand `(3x+5)/((x-1)^2(x+1))` is a proper rational function.

`:.(3x+5)/((x-1)^2(x+1))` `=A/(x-1)+B/((x-1)^2)+C/(x+1)` ---(i)

`=> 3x+5 =A(x-1)(x+1)` `+B(x+1)+C(x-1)^2`

`=A(x^2-1)+B(x+1)` `+C(x^2-2x+1)`

`=>3x+5=(A+C)x^2` `+(B-2C)x+(-A+B+C)`

Now equating the coefficients of like terms on both sides

`A+C=0` -----(ii)

`B-2C=3` -----(iii)

`-A+B+C=5` ----(iv)

By subtracting (iii) from (iv)

`-A+3C=2`

Adding (ii) and (iv)

`C=1/2`

Putting the value of C in (ii)

`A=1/2`

By putting values of C in (iii), we get

`B=4`

Now, by putting values of A, B and C in (i), we get

`(3x+5)/((x-1)^2(x+1))`

`=(-1)/(2(x-1))+4/(x-1)^2+1/(2(x+1))`

`:. int(3x+5)/((x-1)^2(x+1))dx`

`=int[(-1)/(2(x-1))+4/((x-1)^2)+1/(2(x+1))]dx`

`=-1/2int1/(x-1)dx+4int1/(x-1)^2dx` `+1/2int1/(x+1)dx`

`=-1/2log|x-1|+4(x-1)^(-1)/(-1)` `+1/2log|x+1|+C`

`=-1/2log|x-1|-4/(x-1)` `+1/2log|x+1|+C`

`=1/2log|(x+1)/(x-1)|-4/(x-1)+C`

Question: 10. `(2x-3)/((x^2-1)(2x+3))`

Solution:

`int(2x-3)/((x^2-1)(2x+3))dx`

`=int(2x-3)/((x+1)(x-1)(2x+3))dx`

The integrand `(2x-3)/((x+1)(x-1)(2x+3))` is a proper function

`:.(2x-3)/((x+1)(x-1)(2x+3))` `=A/(x+1)+B/(x-1)+C/(2x+3)` ---(i)

`=>2x-3=A(x-1)(2x+3)` `+B(x+1)(2x+3)+C(x+1)(x-1)`

`=>2x-3=A(2x^2+x-3)` `+B(2x^2+5x+3)+C(x^2-1)`

`=>2x-3=(2A+2B+C)x^2` `+(A+5B)x+(-3A+3B-C)`

Eqating coefficients of like terms on both sides, we have

`2A+2B+C=0` ---- (ii)

`A+5B=2` ---- (iii)

`-3A+3B-C=-3` ---- (iv)

`=>3A-3B+C=3`

Subtracting(ii) from (iv), we get

`A-5B=3` ---- (v)

Adding (iii) and (v), we get

`A=5/2`

Putting value of A in (v), we get

`B=-1/10`

Putting values of A and B in (iv), we have

`C=-24/5`

Putting values of A, B and C in (i), we get

 

`(2x-3)/((x+1)(x-1)(2x+3))`

`=5/(2(x+1))-1/(10(x-1))-24/(5(2x+3))`

`:.int(2x-3)/((x+1)(x-1)(2x+3))dx`

`=int[5/(2(x+1))-1/(10(x-1))-24/(5(2x+3))]dx`

`=5/2int1/(x+1)dx-1/10int1/(x-1)dx` `-24/5int1/(2x+3)dx`

`=5/2log|x+1|-1/10log|x-1|` `-24/5(log|2x+3|)/2+C`

`=5/2log|x+1|-1/10log|x-1|` `-12/5log|2x+3|+C`

Question: 11. `(5x)/((x+1)(x^2-4))`

Solution:

`int(5x)/((x+1)(x^2-4))dx`

`=int(5x)/((x+1)(x+2)(x-2))dx`

The integrand `(5x)/((x+1)(x+2)(x-2))` is a proper rational function.

`:.(5x)/((x+1)(x+2)(x-2))`

`=A/(x+1)+B/(x+2)+C/(x-2)` -------(i)

`=>5x=A(x+2)(x-2)` `+B(x+1)(x-2)+C(x+1)(x+2)`

`=>5x=A(x^2-4)` `+B(x^2-x-2)+C(x^2+3x+2)`

`=>5x=(A+B+C)x^2` `+(-B+3C)x+(-4A-2B+2C)`

Equating coefficients of like terms on both sides, we get

`A+B+C=0` -----(ii)

`-B+3C=5` ----(iii)

`-4A-2B+2C=0` ----(iv)

Multiplying (ii) by 4 and then adding with (iv) we get

`B+3C=0` ----(v)

Adding (v) and (v) we get

`C=5/6`

Putting the value of C in (v) we get

`B=-5/2`

Putting the values of B and C in (ii) we get

`A=5/3`

By putting the values of A, B and C in (i) we get

`(5x)/((x+1)(x+2)(x-2))`

`=5/(3(x+1))-5/(2(x+2))+5/(6(x-2))`

`:.int(5x)/((x+1)(x+2)(x-2))dx`

`=int[5/(3(x+1))-5/(2(x+2))+5/(6(x-2))]dx`

`=5/3int1/(x+1)dx-5/2int1/(x+2)dx` `+5/6int1/(x-2)dx`

`=5/3log|x+1|-5/2log|x+2|` `+5/6log|x-2|+C`

Question: 12. `(x^3+x+1)/(x^2-1)`

Solution: `int(x^3+x+1)/(x^2-1)dx`

`=int[x+(2x+1)/(x^2-1)]dx`

`=intxdx+int(2x+1)/(x^2-1)dx`

`=x^2/2+int(2x+1)/(x^2-1)dx`

`=x^2/2+I_1` ------(i)

Now, `I_1=int(2x+1)/(x^2-1)dx`

`=int(2x+1)/((x+1)(x-1))dx`

The integrand `(2x+1)/((x+1)(x-1))` is a proper rational function

`:.(2x+1)/((x+1)(x-1))=A/(x+1)+B/(x-1)`

`=>2x+1=A(x-1)+B(x+1)`

`=>2x+1=(A+B)x+(-A+B)`

Equating coefficients of like terms on both sides, we have

`A+B=2` ----(ii)

`-A+B=1` ----(iii)

Adding (iii) and (v) we get

`B=3/2`

Putting value of B in (iii), we have

`A=1/2`

Putting the value of A and B in (ii), we get

`(2x+1)/((x+1)(x-1))`

`=1/(2(x+1))+3/(2(x-1))`

`:.int(2x+1)/((x+1)(x-1))dx`

`=int[1/(2(x+1))+3/(2(x-1))]dx`

`=1/2int1/(x+1)dx+3/2int1/(x-1)dx`

`=1/2log|x+1|+3/2log|x-1|+C`

Putting the value of `I_1` in (i) we get

`int(x^3+x+1)/(x^2-1)dx`

`=x^2/2+1/2log|x+1|` `+3/2log|x-1|+C`

Question: 13. `2/((1-x)(1+x^2))`

Solution: `int2/((1-x)(1+x^2))dx`

The integrand `2/((1-x)(1+x^2))` is a proper rational function.

`:.2/((1-x)(1+x^2))=A/(1+x)+(Bx+C)/(1+x^2)` ---(i)

`=> 2=A(1+x^2)+(Bx+C)(1-x)`

`=>2=(A+B)x^2+(B-C)x+(A+C)`

Equating coefficients of like terms on both sides

`A-B=0` ---(ii)

`B-C=0` ----(iii)

`A+C=2` ---(iv)

Adding (ii) and (iii), we get

`A-C=0` ---(v)

Adding (iv) and (v), we get

`A=1`

Putting value of A in (ii), we get

`B=1`

Ptting value of A in (iv), we get

`C=1`

Putting values of A, B and C in (i), we get

`2/((1+x)(1+x^2))=1/(1+x)+(x+1)/(1+x^2)`

`:. int2/((1+x)(1+x^2))dx`

`=int[1/(1-x)+(x+1)/(1+x^2)]dx`

`=int1/(1-x)dx+intx/(1+x^2)dx` `+int1/(1+x^2)dx`

`=int1/(1-x)dx+1/2int(2x)/(1+x^2)dx` `+int1/(1+x^2)dx`

`=-log|1-x|+1/2log|1+x^2|` `+tan^-1x+C`

Question: 14. `(3x-1)/(x+2)^2`

Solution: `int(3x-1)/(x+2)^2dx`

The integrand `(3x-1)/(x+2)^2` is a proper rational function.

`:. (3x-1)/(x+2)^2=A/(x+2)+B/(x+2)^2` -----(i)

`=>3x-1=A(x+2)+B`

`=>3x-1=Ax+(2A+B)`

By equating coefficients of like terms on both sides

`A=3` -----(ii)

`2A+B=-1` ----(iii)

Putting value of A in (iii), we get

`B=-7`

Putting values of A, and B inn (i) we get

`(3x-1)/(x+2)^2=3/(x+2)+7/(x+2)^2`

`:. int(3x-1)/(x+2)^2dx`

`=int[3/(x+2)+7/(x+2)^2]dx`

`=3int3/(x+2)dx-7int1/(x+2)^2dx`

`=3 log|x+2|-7(x+2)^(-1)/(-1)+C`

`=3 log|x+2|+7/(x+2)+C`

Question: 15. `1/(x^4-1)`

Solution: `int1/(x^4-1)dx`

=`int1/((x-1)(x+1)(x^2+1))dx`

The integrand `1/((x-1)(x+1)(x^2+1))` is a proper rational function.

`:. 1/((x-1)(x+1)(x^2+1))`

`=A/(x-1)+B/(x+1)+(Cx+D)/(x^2+1)` ----(i)

`=>1=A(x+1)(x^2+1)` `+B(x-1)(x^2+1)` `+(Cx+D)(x-1)(x+1)`

`=>1=A(x^3+x^2+x+1)` `+B(x^3-x^2+x-1)` `+Cx^3+Dx^2-Cx-D`

`=>1=(A+B+C)x^3` `+ (A-B+D)x^2` `+ (A+B-C)x` `+ (A-B-D)`

Equating the coefficients of like terms on both sides, we get

`A+B+C=0` ----(ii)

`A-B+D=0` ----(iii)

`A+B-C=0` -----(iv)

`A-B-D=1` -----(v)

Adding (ii) and (v), we get

`A+B=0` -----(vi)

Adding (iii) and (v), we get

`2A-2B=1` ----(vii)

By multiplying (vi) by 2 and then adding with (vii). we get

`A=1/4`

Putting value of A in (vi), we get

`B=-1/4`

Putting values of A and B in (iv), we get

`C=0`

Putting values of A and B in (v), we get

`D=-1/2`

Putting values of A, B, C and D in (i), we get

`1/((x-1)(x+1)(x^2+1))`

`=1/(4(x-1))-1/(4(x+1))-1/(2(x^2+1))`

`:.int1/((x-1)(x+1)(x^2+1))dx`

`=int[1/(4(x-1))-1/(4(x+1))-1/(2(x^2+1))]dx`

`=1/4int1/(x-1)dx-1/4int1/(x+1)dx` `-1/2int1/(x^2+1)dx`

`=1/4log|x-1|-1/4log|x+1|` `-1/2tan^(-1)x+C`

`=1/4log|(x-1)/(x+1)|-1/2tan^(-1)x+C`

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