# Integral with a substitution

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).

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Is the substitution $x=1/t$ imperative, or could you use a different one? – yunone Jun 23 '11 at 16:43
Unfortunately, it's imperative. – deem Jun 23 '11 at 16:46
Write the integrand as $\displaystyle\frac{t}{\sqrt{t^2 + 1}} = \frac{d}{dt}\sqrt{t^2 + 1}$. – t.b. Jun 23 '11 at 16:47
Sorry, but I don't understand how did you get it? Could you explain it a little more? – deem Jun 23 '11 at 16:53
@user12465: What happens when you divide $1$ by $\frac{a}{b}$? – André Nicolas Jun 23 '11 at 17:19

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

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

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Thank you! I've gotten the same result with the substitution proposed by @Hans Lundmark. – deem Jun 23 '11 at 17:17
@user12465: You are welcome! – Américo Tavares Jun 23 '11 at 17:22

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

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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,

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Why substitute further? you can directly integrate now. – t.b. Jun 23 '11 at 16:44