Question

How do you integrate `int (x-1)/ (x^3 +x)` using partial fractions?

Answer 1

The answer is `=-ln(∣x∣)+1/2ln(x^2+1)+arctanx+C`

Explanation:

Let's factorise the denominator

`x^3+x=x(x^2+1)`

So,

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

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

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

Therefore,

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

Let `x=0`, `=>`, -1=A#

Coefficients of `x^2`

`0=A+B`, `=>`, `B=1`

Coefficients of `x`

`1=C`

So,

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

Therefore,

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

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

We calculate each integral separately

`-intdx/x=-ln(∣x∣)`

Let `u=x^2+1`, `=>`, `du=2xdx`

`int(xdx)/(x^2+1)=int(2du)/u=1/2lnu`

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

Let `x=tantheta`, `=>`, `dx=sec^2theta d theta`

and `x^2+1=tan^2theta + 1=sec^2theta`

So,

`int(1dx)/(x^2+1)=int(sec^2theta d theta)/sec^2theta=intd theta=theta`

`=arctanx`

Putting it alltogether,

`int((x-1)dx)/(x^3+x)=-ln(∣x∣)+1/2ln(x^2+1)+arctanx+C`