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!

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
$$\frac94 \int_1^\infty \frac{dx}{x(x+1)^2}
= \frac94 \lim_{\Lambda\to\infty}\int_1^\Lambda \frac{dx}{x(x+1)^2}
= \frac94 \lim_{\Lambda\to\infty} \left[ \log\frac{x}{x+1} + \frac{1}{1+x}\right]_1^\Lambda
= \frac94 \left( \left(\log 1 + \frac{1}{\infty} \right) - \left(\log\frac12 + \frac12\right)\right)
= \frac98\left(2\log 2 - 1\right)
$$
