Integral with a substitution

Question

I must calculate a following integral

$$\int \frac{dx}{x^{2}\sqrt{1+x^{2}}}$$

with a subsitution like this $x = \frac{1}{t}, t<0$

I'm on this step $$\int \frac{dt}{\frac{1}{t}\sqrt{t^{2} + 1}}$$

I don't know what I should do now... (or maybe it's wrong).

Answer 1

For $t<0$, the substitution $x=1/t$ transforms the integral into

$$\begin{eqnarray*} \int \frac{1}{\left( \frac{1}{t}\right) ^{2}\sqrt{1+\left( \frac{1}{t}% \right) ^{2}}}\left( -\frac{1}{t^{2}}\right) dt &=&-\int \frac{1}{\sqrt{1+% \frac{1}{t^{2}}}}dt \\ &=&-\int \frac{\sqrt{t^{2}}}{\sqrt{t^{2}+1}}dt \\ &=&-\int \frac{\left\vert t\right\vert }{\sqrt{t^{2}+1}}dt=\int \frac{t}{% \sqrt{t^{2}+1}}dt \\ &=&\sqrt{t^{2}+1} \end{eqnarray*}$$

Answer 2

Making the substitution $t=\tan\theta =\frac{\sin \theta}{\cos \theta}$, we see that the integral is $$\int \frac{1}{\tan \theta} d\theta=\int \frac{\cos \theta}{\sin \theta}d\theta.$$ Since $d\left(\sin \theta\right)=-\cos \theta$ we can write this antiderivative as $-\ln \sin \theta $.

Hope that helps,

Answer 3

If you don't see right away how to integrate it, try substituting $s=t^2+1$.