1
$\begingroup$

I have a function that I am trying to split into partial fractions in order to integrate the function.

The function is:

$$\int \frac{x}{(1+x^2)^2}dx$$

I am trying to split $\dfrac{x}{(1+x^2)^2}$ into partial fractions.

While trying I am splitting it into these fractions: $\dfrac{Ax+D}{1+x^2}+\dfrac{Cx+D}{(1+x^2)^2}$

I get $C=1$ which ends up again where I started from. I am confused on what to do here.

  • 1
    What you have discovered is that partial fractions is not a useful technique for this particular problem. – 2017-02-13

2 Answers 2

2

Hint. Alternatively, one may just perform the change of variable $$ u=1+x^2, \quad \frac12du=x\:dx, \quad $$ giving $$ \int \frac{x\:dx}{(1+x^2)^2}=\frac12\int \frac{du}{u^2} $$ the latter integral being easier to evaluate.

  • 0
    This is what you are looking for, this solves as $\frac {-1}{2+2x^2} + C$ – 2017-02-13
1

Since $1+x^2 =(1+ix)(1-ix) $, if we ask Wolfy for $\dfrac{1}{(1+ax)^2(1+bx)^2} $ we get $-\dfrac{2 a^2 b}{(a - b)^3 (a x + 1)} + \dfrac{a^2}{(a - b)^2 (a x + 1)^2} + \dfrac{2 a b^2}{(a - b)^3 (b x + 1)} + \dfrac{b^2}{(a - b)^2 (b x + 1)^2} $.

Putting $a=i, b=-i$, we get both

$\dfrac{i}{4 (x + i)} - \dfrac{1}{4 (x + i)^2} - \dfrac{i}{4 (x - i)} - \dfrac{1}{4 (x - i)^2} $

and

$\dfrac{2 x^2}{(x^2 + 1)^4} + \dfrac{1}{(x^2 + 1)^4} + \dfrac{x^4}{(x^2 + 1)^4} $.