How to evaluate this particular improper intergral I am studying for upcoming exams and I am wondering how I would evaluate this question? (I have used a calculator and gotten an answer of 0.43, this answer is acceptable for me but I do not know how to get to there manually) 
$$\int_1^\infty\dfrac{18}{8x(x+1)^2}\,dx$$
Thanks!
 A: Hint. You may use a partial fraction decomposition to get
$$
\dfrac{18}{8x(x+1)^2}=\frac{9}{4x}-\frac{9}{4 (1+x)}-\frac{9}{4 (1+x)^2}
$$ then integrate each term on the finite interval $[1,M]$ ($M\geq1$)
$$
\int_1^M\dfrac{18}{8x(x+1)^2}dx=\int_1^M\frac{9}{4x}dx-\int_1^M\frac{9}{4 (1+x)}dx-\int_1^M\frac{9}{4 (1+x)^2}dx
$$ then let $M \to +\infty$.
You will find 

$$ \int_1^{\infty}\dfrac{18}{8x(x+1)^2}dx=\frac{9 \ln 2}{4}-\frac{9}{8}=\color{blue}{0.434581156259\cdots}.$$

A: The canonical way is to perform partial fraction decomposition. For this particular integrand, one trick that is useful is
$$\frac{1}{x(x+a)} = \frac{1}{a}\left(\frac{1}{x} - \frac{1}{x+a}\right)$$
With this, you can basically read off the partial fraction decomposition of the integrand as:
$$\frac{1}{x(x+1)^2} = \left(\frac{1}{x} - \frac{1}{x+1}\right)\frac{1}{x+1} = 
\frac{1}{x(x+1)} - \frac{1}{(x+1)^2} = \frac{1}{x} - \frac{1}{x+1} - \frac{1}{(x+1)^2}$$
So the integral becomes
$$\frac94 \int_1^\infty \frac{dx}{x(x+1)^2}
= \frac94 \lim_{\Lambda\to\infty}\int_1^\Lambda \frac{dx}{x(x+1)^2}
= \frac94 \lim_{\Lambda\to\infty} \left[ \log\frac{x}{x+1} + \frac{1}{1+x}\right]_1^\Lambda
= \frac94 \left( \left(\log 1 + \frac{1}{\infty} \right) - \left(\log\frac12 + \frac12\right)\right)
= \frac98\left(2\log 2 - 1\right)
$$
A: Transform the integral to the interval $[0,\infty]$ and use the residue theorem, i.e.,
$$\int_0^{\infty} \frac{dx}{(x+1)(x+2)^2} $$
Then consider the integral 
$$\oint_C dz \frac{\log{z}}{(z+1)(z+2)^2}$$
where $C$ is a keyhole contour of outer radius $R$ and inner radius $\epsilon$ about the positive real axis.  Note that we define the branch of the log by confining $\arg{z} \in [0,2 \pi)$.  
As $R \to \infty$ and $\epsilon \to 0$, the contour integral approaches
$$-i 2 \pi \int_0^{\infty} \frac{dx}{(x+1)(x+2)^2} $$
and, by the residue theorem, the contour integral is equal to $i 2 \pi$ times the sum of the residues at the poles $z=e^{i \pi}$ and $z=2 e^{i \pi}$, or
$$i 2 \pi \left (\frac{i \pi}{(2-1)^2} + \frac1{(-2)(-2+1)} - \frac{\log{2}+i \pi}{(-2+1)^2}\right ) = i 2 \pi \left (\frac12 - \log{2} \right )$$
Thus,
$$\frac{9}{4} \int_0^{\infty} \frac{dx}{(x+1)(x+2)^2} = \frac{9}{8} (2 \log{2}-1)$$
