Partial Fraction Decomposition: $\int_{0}^\infty \frac{xdx}{(x+\theta)^4}$

This is a simple calculus question, but it's been a while since I've had to use partial fraction decomposition. I need to integrate the following:

$$\int_{0}^\infty \frac{xdx}{(x+\theta)^4}$$

I approached this via partial fractions and used the following decomposition:

$$\frac{x}{(x+\theta)^4} = \frac{A}{x+\theta} + \frac{B}{(x+\theta)^2} +\frac{C}{(x+\theta)^3} + \frac{D}{(x+\theta)^4}$$

which led to

$$x = (A)x^3 + (A\theta +2A\theta +B)x^2 + (A2\theta^2 + A \theta^2 +2\theta +C)x + (B\theta^2 + D + A\theta^3)$$

then equating coefficients yields the following system

$$A = 0 \\ A\theta +2A\theta + B = 0 \\ 2A\theta^2 + A \theta^2 + 2\theta + C = 1 \\ B\theta^2 + D + A\theta^3 = 0$$

resulting in the following values:

$$A=0\\ B=0\\ C=1-2\theta\\ D=0$$

and plugging these back in gives

$$\frac{x}{(x+\theta)^4} = \frac{1-2\theta}{(x+\theta)^3}$$

But, Wolfram is telling me that

$$\frac{x}{(x+\theta)^4} = \frac{1}{(x+\theta)^3} - \frac{\theta}{(x+\theta)^4}.$$

I thought that maybe these would yield the same result, but they definitely do not. I've double-checked my work many times here and can't find my mistake. If anyone could help me out here, I'd appreciate. Or, of course, if anyone knows a simpler way to tackle this integral, that'd obviously be appreciated, as well. Thanks!

EDIT: Yes, of course, a simple substitution suffices here. Sheesh... that's a little embarrassing. Thanks, everyone. Still, I am curious as to what went wrong with my partial fraction decomposition...

• Check you work again. – David C. Ullrich Aug 8 '15 at 16:18
• The last two terms in your third equation are wrong. You should get $(3A\theta^2+2B\theta+C)x+(A\theta^3+B\theta^2+C\theta+D)$ instead. This way you get exactly Wolfram's result when comparing coefficients. – Arnaud D. Aug 8 '15 at 16:25

HINT:

A better way will be to replace $x+\theta$ with $y$

$x\to\infty\implies y\to\infty, x=0\implies y=\theta$

Hint: $\int_0^{\infty} \frac{xdx}{(x+\theta )^k} =\int_{\theta }^{\infty} \frac{u-\theta}{u^k} du$

we have $$\int_{0}^\infty \frac{xdx}{(x+\theta)^4}$$ Let, $x+\theta=t\implies dx=dt$

$$\int_{\theta}^\infty \frac{(t-\theta)dt}{t^4}$$ $$\int_{\theta}^\infty (t^{-3}-\theta t^{-4})dt$$ Can you take it from here?

Just observe \begin{align*} \frac{x}{(x+\theta)^4}&=\frac{x+\theta-\theta}{(x+\theta)^4}\\ &=\frac{x+\theta}{(x+\theta)^4}-\frac{\theta}{(x+\theta)^4}\\ &=\frac{1}{(x+\theta)^3}-\frac{\theta}{(x+\theta)^4} \end{align*}