Question

Find:

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

Answer 1

Let

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

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

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

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

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

`=x+C_1-int(x+1)/(x^3+x)dx`

`then I=x+c_1+I_1...................(i)`

now

`I_1=int(x+1)/(x^3+x)dx`

`=>I_1=int(x+1)/(x(x^2+1))dx`

`Let (x+1)/(x(x^2+1))=A/x+(Bx+C)/(x^2+1)`

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

Comparing the coefficients of numerator, we get
A = 1, B = − 1 and C = 1

`So I_1=int(x+1)/(x(x^2+1))dx=int1/x dx+int(-x+1)/(x^2+1)dx`

`=>I_1=log|x|+int(-x+1)/(x^2+1)dx`

`=>I_1=log|x|-1/2int(2x)/(x^2+1)dx+int1/(x^2+1)dx`

`=>I_1=log|x|-1/2log|x^2+1|+tan^(-1)(x^2+1)+C_2..................(ii)`

From (i) and (ii), we get

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