# How to integrate $\int {2\over (x^2+2)\sqrt{x^2+4}}dx$?

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 $$x$$, $$=\tan^{-1}\left(\frac{x}{\sqrt{x^2+4}}\right)+C$$

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 $$\int - \frac{du}{u^2 + 1}$$ $$=- \tan^{-1} u = -\tan^{-1}(\sqrt{1+4t^2}) = -\tan^{-1}\sqrt{1 + \frac{4}{x^2}}$$ $$= \tan^{-1}\frac{x}{\sqrt{x^2+4}} + C$$

• thank you for answer – user809080 Jul 28 '20 at 21:24

Alternatively, let $$x = 2\sinh t \Rightarrow \frac{dx}{dt} = 2\cosh t$$.

The integral becomes

$$\begin{array} {r c l } \displaystyle \int \frac2{(4\sinh^2 t + 2)(2\cosh t)} \cdot 2\cosh t \, dt &=& \displaystyle \frac12 \int \frac1{2\sinh^2 t + 1} \, dt \\ &=& \displaystyle \frac12 \tan^{-1} (\tanh t) + C \\ &=& \displaystyle \frac12 \tan^{-1} \left (\tanh \left (\sinh^{-1} \tfrac x2 \right )\right) + C \\ \end{array}$$

• thank you for answer – user809080 Jul 28 '20 at 21:24