Ali's Improper Integrals: Problem 1

as suggested by @Sudeep Salgia $\begin{array}{c}(\dfrac{2x^2-1}{x^4+5x^2+4}\\ =\dfrac{3x^2+3-x^2-4}{x^4+4x^2+x^2+3}\\ =\dfrac{3(x^2+1)-(x^2+4)}{(x^2+4)(x^2+1)}\\ =\dfrac{3}{x^2+4}-\dfrac{1}{x^2+1}\end{array}$ hence $\int_0^{\infty}\frac{ 3(x^2 + 1) - (x^2 + 4)}{(x^2+1)(x^2+4)} dx= 3 \int_0^{\infty} \frac{1}{x^2 +4} dx - \int_0^{\infty} \frac{1}{x^2 +1} dx$ lets solve each integral. $\int \frac{1}{x^2 +4} dx$ let $\begin{cases} u=arctan(2x)\\x= 2tan(u)\\ dx=2sec^2(x) du\end{cases}$ so: $\int \frac{1}{x^2 +4} dx=\int \frac{1}{4tan^2(x) +4} 2(sec^2(x)) du$ by a fundemental trig identity: $=\int \frac{1}{4sec^2(x)} 2(sec^2(x)) du=\int \frac{1}{2} du$ $=\dfrac{1}{2}u+C=\dfrac{arctan(2x)}{2}+C$ great, second part: $\int \frac{1}{x^2 +1}dx$ so: $\begin{cases} u=arctan(x)\\x=tan(u)\\dx=sec^2(u) du\end{cases}$ put this in $\int \frac{1}{x^2 +1}dx=\int \dfrac{1}{tan^2(u)+1}sec^2(u)du=\int \dfrac{1}{sec^2(u)}sec^2(u)du =\int du=u+C=arctan(x)+C$ so we are looking for $lim_{n\rightarrow\infty} [(3(\dfrac{arctan(2x)}{2})-arctan(x)]_0^n=lim_{n\rightarrow\infty} (3(\dfrac{arctan(2n)}{2})-arctan(n)-3*0+0)$ $=3\dfrac{\dfrac{\pi}{2}}{2}-\dfrac{\pi}{2}=\boxed{\dfrac{\pi}{4}}$

EDIT: This answer uses the tools of complex analysis to get a result, Sudeep Salgia has a much simpler solution as a comment, please upvote his solution instead!

Let $I=\int_0^\infty \frac{2x^2-1}{x^4+5x^2+4}dx=\frac12\int_{-\infty}^\infty\frac{2x^2-1}{(x^2+1)(x^2+4)}dx$ The poles of the integrand are at $i$ and $2i$ , we take the upper semicircle along the real axis, centred at $0$ of radius $R$ as our contour. Hence we have that $\oint_C\frac{2z^2-1}{(z^2+1)(z^2+4)}dz=\int_{C_1}\frac{2z^2-1}{(z^2+1)(z^2+4)}dz+\underbrace{\int_{-R}^R\frac{2x^2-1}{(x^2+1)(x^2+4)}dx}_{2I\ \text{as}\ R\to\infty}$ where $C_1$ is the circular part of the semicircle contour. We evaluate the integral covering $C_1$ : Note that $C_1$ is given by $z=Re^{it}$ for $t\in[0,\pi]$ , then $dz=iRe^{it}dt$ so: $\int_{C_1}\frac{2z^2-1}{(z^2+1)(z^2+4)}dz=\int_{0}^\pi\frac{iR(2{Re^{it}}^2-1)}{((Re^{it})^2+1)((Re^{it})^2+4)} dt\\=\int_{0}^\pi\frac{iR^2(2{e^{it}}^2-1/R)}{R^4((e^{it})^2+1/R^2)((e^{it})^2+4/R^2)} dt$ which clearly goes to $0$ as $R\to\infty$ . Now by the Residue Theorem: $\oint_C\frac{2z^2-1}{(z^2+1)(z^2+4)}dz=2\pi i (b_{i}+b_{2i})$ Where $b_1$ and $b_{2}$ are the residues at $i$ and $2i$ respectively. These are simple poles of $f(z)=\frac{2z^2-1}{(z^2+1)(z^2+4)}$ which can be written as $f(z)=\frac{\phi_1(z)}{z-i}=\frac{\phi_2(z)}{z-2i}$ so $b_i=\phi_1(i)=i/2$ and $b_{2i}=\phi_2(2i)=-3i/4$ hence: $\oint_C\frac{2z^2-1}{(z^2+1)(z^2+4)}dz=2\pi i (b_{i}+b_{2i})=\pi/2$ and so as $R\to \infty$ , $2I=\pi/2\implies \boxed{I=\pi/4}$

Can't we write 2-1\x^(2)= m(1-2\x^(2))+n(1+2\x^(2)) derivatives of x+2\x and x-2/x I did by this way ... Edit-this trick has been done after multiplying and dividing the integrand by x^(2)