Evaluate:
$u =\displaystyle\int_0^\infty\dfrac{dx}{x^4 +7x^2+1}$
Attempt:
$$u = \int_0^\infty \dfrac{dx}{\left(x^2+ \left(\dfrac{7 - \sqrt {45}}{2}\right)\right)\left(x^2+ \left(\dfrac{7 + \sqrt {45}}{2}\right)\right)}$$
$$u = \int_0^\infty \dfrac{dx}{(x^2+a^2)(x^2+b^2)}$$
After partial fraction decomposition and simplifying I get:
$u = \dfrac{\pi}{2(a+b)}$
But answer is $\frac \pi 6$.
Where have I gone wrong?
On
Hint:
$$\dfrac2{x^4+ax^2+1}=\dfrac{1-1/x^2}{x^2+a+1/x^2}+\dfrac{1+1/x^2}{x^2+a+1/x^2}$$
$$x^2+1/x^2=(x-1/x)^2+?=(x+1/x)^2-2$$
On
I'm not sure you have done anything wrong. You have $a+b$ in your answer, but $a$ is specifically $\sqrt{\frac{7-\sqrt{45}}{2}}=\frac{3-\sqrt{5}}{2}$. And $b=\frac{3+\sqrt{5}}{2}$. So $a+b=3$, making your answer agree with what you are expecting.
On
Notice that the fraction could be written as
\begin{equation}
{\displaystyle\int}\dfrac{1}{\left(x^2-\frac{3\cdot\sqrt{5}-7}{2}\right)\left(x^2+\frac{3\cdot\sqrt{5}+7}{2}\right)}\,\mathrm{d}x
\end{equation}
After performing partial fraction decomposition, we get
\begin{equation}
{\displaystyle\int}\left(\dfrac{2}{3\cdot\sqrt{5}\left(2x^2-3\cdot\sqrt{5}+7\right)}-\dfrac{2}{3\cdot\sqrt{5}\left(2x^2+3\cdot\sqrt{5}+7\right)}\right)\mathrm{d}x
\end{equation}
Let's solve
\begin{equation}
{\displaystyle\int}\dfrac{1}{2x^2-3\cdot\sqrt{5}+7}\,\mathrm{d}x
\end{equation}
Use the change of variable
\begin{equation}
u=\dfrac{\sqrt{2}x}{\sqrt{7-3\cdot\sqrt{5}}}
\end{equation}
This gives
\begin{equation}
\mathrm{d}x=\dfrac{\sqrt{7-3\cdot\sqrt{5}}}{\sqrt{2}}\,\mathrm{d}u
\end{equation}
The integral in the third equation becomes
\begin{equation}
={\displaystyle\int}\dfrac{\sqrt{7-3\cdot\sqrt{5}}}{\sqrt{2}\left(\left(7-3\cdot\sqrt{5}\right)u^2-3\cdot\sqrt{5}+7\right)}\,\mathrm{d}u
\end{equation}
It simplifies to
\begin{equation}
{{\dfrac{1}{\sqrt{2}\sqrt{7-3\cdot\sqrt{5}}}}}{\displaystyle\int}\dfrac{1}{u^2+1}\,\mathrm{d}u
\end{equation}
The integral above evaluates as
\begin{equation}
{{\dfrac{1}{\sqrt{2}\sqrt{7-3\cdot\sqrt{5}}}}}{\displaystyle\int}\dfrac{1}{u^2+1}\,\mathrm{d}u
=
\dfrac{\arctan\left(u\right)}{\sqrt{2}\sqrt{7-3\cdot\sqrt{5}}}
\end{equation}
Using $u$ in the fourth equation gives
\begin{equation}
{{\dfrac{1}{\sqrt{2}\sqrt{7-3\cdot\sqrt{5}}}}}{\displaystyle\int}\dfrac{1}{u^2+1}\,\mathrm{d}u
=
\dfrac{\arctan\left(u\right)}{\sqrt{2}\sqrt{7-3\cdot\sqrt{5}}}
=
\dfrac{\arctan\left(\frac{\sqrt{2}x}{\sqrt{7-3\cdot\sqrt{5}}}\right)}{\sqrt{2}\sqrt{7-3\cdot\sqrt{5}}}
\end{equation}
Now let's solve the second integral appearing in the second equation, i.e.
\begin{equation}
{\displaystyle\int}\dfrac{1}{2x^2+3\cdot\sqrt{5}+7}\,\mathrm{d}x
\end{equation}
Change of variable as
\begin{equation}
u=\dfrac{\sqrt{2}x}{\sqrt{3\cdot\sqrt{5}+7}}
\end{equation}
i.e.
\begin{equation}
\mathrm{d}x=\dfrac{\sqrt{3\cdot\sqrt{5}+7}}{\sqrt{2}}\,\mathrm{d}u
\end{equation}
We get
\begin{equation}
{\displaystyle\int}\dfrac{\sqrt{3\cdot\sqrt{5}+7}}{\sqrt{2}\left(\left(3\cdot\sqrt{5}+7\right)u^2+3\cdot\sqrt{5}+7\right)}\,\mathrm{d}u
\end{equation}
which is
\begin{equation}
{\displaystyle\int}\dfrac{\sqrt{3\cdot\sqrt{5}+7}}{\sqrt{2}\left(\left(3\cdot\sqrt{5}+7\right)u^2+3\cdot\sqrt{5}+7\right)}\,\mathrm{d}u
=
{{\dfrac{1}{\sqrt{2}\sqrt{3\cdot\sqrt{5}+7}}}}{\displaystyle\int}\dfrac{1}{u^2+1}\,\mathrm{d}u
=\dfrac{\arctan\left(u\right)}{\sqrt{2}\sqrt{3\cdot\sqrt{5}+7}}
\end{equation}
Use $u$ in the $11^{th}$ equation, i.e.
\begin{equation}
{\displaystyle\int}\dfrac{\sqrt{3\cdot\sqrt{5}+7}}{\sqrt{2}\left(\left(3\cdot\sqrt{5}+7\right)u^2+3\cdot\sqrt{5}+7\right)}\,\mathrm{d}u
=
{{\dfrac{1}{\sqrt{2}\sqrt{3\cdot\sqrt{5}+7}}}}{\displaystyle\int}\dfrac{1}{u^2+1}\,\mathrm{d}u
=\dfrac{\arctan\left(\frac{\sqrt{2}x}{\sqrt{3\cdot\sqrt{5}+7}}\right)}{\sqrt{2}\sqrt{3\cdot\sqrt{5}+7}}
\end{equation}
Substituting the results we have in the second equation which is
\begin{equation}
{{\dfrac{2}{3\cdot\sqrt{5}}}}{\displaystyle\int}\dfrac{1}{2x^2-3\cdot\sqrt{5}+7}\,\mathrm{d}x-{{\dfrac{2}{3\cdot\sqrt{5}}}}{\displaystyle\int}\dfrac{1}{2x^2+3\cdot\sqrt{5}+7}\,\mathrm{d}x
\end{equation}
gives
\begin{equation}
\dfrac{\sqrt{2}\arctan\left(\frac{\sqrt{2}x}{\sqrt{7-3\cdot\sqrt{5}}}\right)}{3\cdot\sqrt{5}\sqrt{7-3\cdot\sqrt{5}}}-\dfrac{\sqrt{2}\arctan\left(\frac{\sqrt{2}x}{\sqrt{3\cdot\sqrt{5}+7}}\right)}{3\cdot\sqrt{5}\sqrt{3\cdot\sqrt{5}+7}}
\end{equation}
Evaluating at the limits gives:
\begin{equation}
-\dfrac{\sqrt{3\cdot\sqrt{5}+7}\left(7\cdot\sqrt{2}\cdot\sqrt{5}-15\cdot\sqrt{2}\right){\pi}+\sqrt{7-3\cdot\sqrt{5}}\left(-7\cdot\sqrt{2}\cdot\sqrt{5}-15\cdot\sqrt{2}\right){\pi}}{120}
=
\frac{\pi}{6}
\end{equation}
On
A slightly different approach: you have $a^2+b^2=7$ and $a^2 b^2=1$ with $a,b\in\mathbb{R}^+$, such that
$$ \operatorname*{Res}_{x=ia}\frac{1}{(x^2+a^2)(x^2+b^2)} = \lim_{x\to ia}\frac{1}{(x+ia)(x^2+b^2)} = \frac{1}{2i}\cdot \frac{1}{a(b^2-a^2)}$$ and similarly $$ \operatorname*{Res}_{x=ib}\frac{1}{(x^2+a^2)(x^2+b^2)} = \lim_{x\to ib}\frac{1}{(x+ib)(x^2+a^2)} = \frac{1}{2i}\cdot \frac{1}{b(a^2-b^2)}$$ so $\pi i$ times the sum of the residues at $ia$ and $ib$ equals $\frac{\pi}{2ab(a+b)}$. On the other hand the integral is blatantly positive, such that $ab(a+b)$ is the square root of $a^2 b^2(a^2+b^2+2\sqrt{a^2 b^2}) = 9$ and voilà the wanted outcome $\frac{\pi}{6}$.
$\newcommand{\I}{\mathfrak{I}}$$\newcommand{\dx}{\mathrm dx\,}$Let’s take the time to generalize this. Denote the generalized integral as$$\I=\int\limits_0^{\infty}\dx\frac 1{x^4+ax^2+b^2}$$Now factor out an $x^2$ from the denominator and complete the square to get $$\I=\int\limits_0^{\infty}\frac {\dx}{x^2}\frac 1{\left(x-\frac bx\right)^2+a+2b}$$Make the transformation $x\mapsto\tfrac bx$ so that$$\I=\frac 1b\int\limits_0^{\infty}\dx\frac 1{\left(x-\frac bx\right)^2+a+2b}$$Therefore, it’s easy to see that $$b\I=\int\limits_0^{\infty}\frac {\dx}{x^2}\frac b{\left(x-\frac bx\right)^2+a+2b}=\int\limits_0^{\infty}\dx\frac 1{\left(x-\frac bx\right)^2+a+2b}$$ Add the two integrals together to get $$\begin{align*}\I & =\frac 1{2b}\int\limits_0^{\infty}\dx\left(1+\frac b{x^2}\right)\frac 1{\left(x-\frac bx\right)^2+a+2b}\\ & =\frac 1{2b}\int\limits_{-\infty}^{\infty}\frac {\dx}{x^2+a+2b}\\ & \color{blue}{=\frac {\pi}{2b\sqrt{a+2b}}}\end{align*}$$ Now set $a=7$ and $b=1$ to get the answer.