Solve the following indefinite integral: $$\int \frac{2}{(x^2+2)\sqrt{x^2+4}} dx$$
My approach:
I used the substitution: $x=2\tan t$, $dx=2\sec^2t dt$
$$\int \frac{2}{(x^2+2)\sqrt{x^2+4}} dx=\int \frac{2}{(4\tan^2t+2)\sqrt{4\tan^2t+4}}\cdot 2\sec^2t\ dt$$
$$=\int \frac{4\sec^2t }{2(2\tan^2t+1)2\sec t} dt$$ $$=\int \frac{\sec t}{2\tan^2t+1}dt$$
In numerator I have $\sec t$ but not $\sec^2t$ therefore I can't see a way to take it further. Please help me solve this integral. Thanks in advance.
Your substitution is correct. Continue as follows $$=\int \frac{\sec t\ dt }{ 2\tan^2t+1}$$ $$=\int \frac{\frac{1}{\cos t}\ dt }{ 2\frac{\sin^2t}{\cos^2t}+1}$$ $$=\int \frac{\cos t\ dt }{ 2\sin^2t+\cos^2t}$$ $$=\int \frac{d(\sin t) }{ \sin^2t+1}$$ $$=\tan^{-1}(\sin t)+C$$ substituting back to $x$, $$=\tan^{-1}\left(\frac{x}{\sqrt{x^2+4}}\right)+C$$
To remove the square root, let $x=\frac{2t}{\sqrt{1-t^2}} $, or $t=\frac x{\sqrt{x^2+4}}$ $$\int \frac{2}{(x^2+2)\sqrt{x^2+4}}dx=\int\frac{1}{1+t^2}dt =\tan^{-1}t $$
The problem's already solved but just another method.
Substitute $x = \frac{1}{t}$
$$=\int \frac{2}{\big(\frac{1}{t^2} + 2\big) \sqrt{\frac{1}{t^2} + 4}} \frac{-dt}{t^2}$$
$$=\int \frac{2t^3}{(1+2t^2) \sqrt{1+4t^2}} \frac{-dt}{t^2}$$
Now substitute $1+4t^2 = u^2$.
Integral will simplify to $$\int - \frac{du}{u^2 + 1}$$ $$=- \tan^{-1} u = -\tan^{-1}(\sqrt{1+4t^2}) = -\tan^{-1}\sqrt{1 + \frac{4}{x^2}}$$ $$ = \tan^{-1}\frac{x}{\sqrt{x^2+4}} + C$$
Alternatively, let $ x = 2\sinh t \Rightarrow \frac{dx}{dt} = 2\cosh t$.
The integral becomes
$$\begin{array} {r c l } \displaystyle \int \frac2{(4\sinh^2 t + 2)(2\cosh t)} \cdot 2\cosh t \, dt &=& \displaystyle \frac12 \int \frac1{2\sinh^2 t + 1} \, dt \\ &=& \displaystyle \frac12 \tan^{-1} (\tanh t) + C \\ &=& \displaystyle \frac12 \tan^{-1} \left (\tanh \left (\sinh^{-1} \tfrac x2 \right )\right) + C \\ \end{array} $$
Make everything in terms of $\sec t$.
$\int\frac{\sec t}{2\tan^2t+1}dt=\int\frac{\sec t}{2\left(\sec^2t-1\right)+1}dt$
$=\int\frac{\sec t}{2\sec^2t-1}dt$
Multiply the numerator and denominator by $\cos^2t$.
$\int\frac{\sec t}{2\sec^2t-1}dt=\int\frac{\sec t}{2\sec^2t-1}\times\frac{\cos^2t}{\cos^2t}dt$
$=\int\frac{\cos t}{2-\cos^2t}dt$
Rewrite $\cos^2t$ as $1-\sin^2t$.
$\int\frac{\cos t}{2-\cos^2t}dt=\int\frac{\cos t}{2-\left(1-\sin^2t\right)}dt$
$=\int\frac{\cos t}{1+\sin^2t}dt$
Use the substitution $y=\sin t=\frac{x}{\sqrt{x^2+4}}$ and $dy=\cos t\ dt$.
$\int\frac{\cos t}{1+\sin^2t}dt=\int\frac{1}{1+y^2}dy$
$=\tan^{-1}y+c$
$=\tan^{-1}\left(\frac{x}{\sqrt{x^2+4}}\right)+c$
