How resolve this integral?

Question

How resolve? $$ \int \frac 1 {x^2\sqrt{x^2+4} } dx $$ I try

$x=2\tan \theta$ $$ \int \frac 1 {4\tan ^2 \theta \sqrt{4\tan ^2 \theta +4} } dx= \int \frac 1 {4\tan ^2 \theta \sqrt{4(\tan^2 \theta +1) } }dx $$ $\tan^2\theta +1=\sec^2\theta$ $$ \int \frac 1 {4\tan ^2\theta \sqrt {4\sec ^2\theta} } dx $$

Answer 1

You are on the right track.

You applied $u$ substitution $x=2\tan\theta$ and $dx=2\sec^2\theta\ d\theta$

Then you get $$\int\dfrac{\sec\theta}{4\tan^2\theta}\ d\theta=\dfrac14\int\dfrac{\dfrac{1}{\cos\theta}}{\left(\dfrac{\sin\theta}{\cos\theta}\right)^2}=\dfrac14\int\dfrac{\cos\theta}{\sin^2\theta}$$ Now again apply substitution $v=\sin\theta$ $$=\dfrac14\int\dfrac{1}{v^2}\ dv=-\dfrac14\cdot\dfrac{1}{v}$$

Now substitute back $v=\sin\theta$ and so on......

Can you take it from here?

Answer 2

Another substitution:

You can set $\; x=2\sinh t$, $\;\mathrm d x=\cosh t\,\mathrm dt$ This is bijective substitution, and we obtain the integral: $$\int \frac 1 {x^2\sqrt{x^2+4}}\,\mathrm d x=\int\frac{\cosh t\,\mathrm dt} {\sinh^2t\cdot2\sqrt{\sinh^2t+1}}=\frac12\int\frac{\mathrm dt} {\sinh^2t}=-\frac12\coth t. $$

Now, $\;t=\operatorname{argsinh}\bigl(\frac x2\bigr)$, so $\;\sinh t=\frac x2$, $\cosh t=\dots$

Answer 3

Here is a "non-trigonometric" way:

• Substitute $x = \frac{1}{y}\Rightarrow dx = -\frac{1}{y^2}dy$ $$\int \frac 1 {x^2\sqrt{x^2+4} } dx = -\int \frac{y^2}{\sqrt{\frac{1}{y^2}+4}}\cdot \frac{1}{y^2}\; dy \stackrel{y>0}{=} -\int \frac{y}{\sqrt{1+4y^2}}\;dy$$ $$= -\frac{1}{4}\sqrt{1+4y^2} = -\frac{1}{4}\sqrt{1+4\frac{1}{x^2}} = -\frac{\sqrt{x^2+4}}{4x}(+C)$$

Deriving the expression shows that this is also the antiderivative in the case $y< 0 \Leftrightarrow x<0$.