Question

${\int }_0^{\infty }\frac{dt}{(1+t^2)(1+4t^2)}$

Answer 1

$\begin{array}{c}{\int }_0^{\infty }\frac{dt}{\left(1+t^2\right)\left(1+4t^2\right)}\\ put{t}^2=y\\ \Rightarrow \frac{1}{\left(1+y\right)\left(1+4y\right)}=\frac{A}{\left(1+y\right)}+\frac{B}{\left(1+4y\right)}\\ \Rightarrow \frac{1}{\left(1+y\right)\left(1+4y\right)}=\frac{A+4AY+B+BY}{\left(1+y\right)\left(1+4y\right)}\\ A+B=1\\ -4A=B\\ PutB=-4Ainequation\left(i\right)\\ \Rightarrow A-4A=1\\ A=\frac{-1}{3}\\ andB=\frac{4}{3}\\ \Rightarrow {\int }_0^{\infty }\frac{dt}{\left(1+t^2\right)\left(1+4t^2\right)}=\frac{-1}{3}{\int }_0^{\infty }\frac{1}{1+t^2}dt+\frac{4}{3}{\int }_0^{\infty }\frac{dt}{1+4t^2}\\ \Rightarrow \frac{-1}{3}{\int }_0^{\infty }\frac{1}{\left(1+t^2\right)}dt+\frac{1}{3}{\int }_0^{\infty }\frac{dt}{\left[{\left(\frac{1}{2}\right)}^2+t^2\right]}\\ =\left[\frac{-1}{3}{\tan }^{-1}t+\frac{1}{3}\times 2{\tan }^{-1}2t\right]+c\\ {\int }_0^{\infty }\frac{dt}{\left(1+t^2\right)\left(1+4t^2\right)=\frac{1}{3}}\left[2{\tan }^{-1}2t-{\tan }^{-1}\right]+c\\ =\frac{1}{3}\left[2{\tan }^{-1}\infty -{\tan }^{-1}\infty -0\right]+c\\ =\frac{1}{3}\left[\pi -\frac{\pi }{2}-0\right]=\frac{\pi }{6}\end{array}$