How do you integrate `int (3x) / (x^2 * (x^2+1) )` using partial fractions?

The expression needs some simplification

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

Now we can perform the decomposition into partial fractions

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

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

The denominators are the same, we compare the numerators

`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`,

`0=C`

Therefore,

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

So,

the integral is

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

The second integral is done by substutution

Let `u=x^2+1`

`du=2xdx`

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

`=1/2lnu`

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

Putting it all together,

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

Resprcted Narad T. has solved the Problem using Partial

Fractions, as is reqd. to be done in that fashion.

As an Aliter, I submit the following Soln. :

Let `I=int(3x)/{x^2(x^2+1)}dx=3/2int(2x)/{x^2(x^2+1)}dx.`

Noting that, `d/dx(x^2)=2x,` we subst. `x^2=t," so, "2xdx=dt.`

`:. I=3/2int1/{t(t+1)}dt=3/2int{(t+1)-(t)}/{t(t+1)}dt,`

`=3/2int{(t+1)/(t(t+1))-t/(t(t+1))}dt,`

`=3/2int{1/t-1/(t+1)}dt,`

`=3/2[ln|t|-ln|t+1|},`

`=3/2ln|t/(t+1)|, and, because, t=x^2,`

`I=3/2ln{x^2/(x^2+1)}+C, or,`

`I=3/2lnx^2-3/2ln(x^2+1)+C=3ln|x|-3/2ln(x^2+1)+C.`

Enjoy Maths.!