Breaking up an integral using substitution? I am faced with the following integral, and am required to use substitution to solve it, using $u = \frac{1}{x}$. 
$$\int \dfrac{1}{x^2\sqrt{x^2+1}} \, dx$$
I know I can break up the equation in:
$$\int \dfrac{1}{x}\dfrac{1}{x}\dfrac{1}{\sqrt{x^2+1}} \, dx.$$
How would I go about replacing the variable $x$ in the sqrt though? 
 A: I suggest you to use $$\frac{\sqrt{1+x^2}}{x}=t$$ So you 'll find the integral reduced to the following one: $$\int -dt$$
A: \begin{align}
u & = \frac1x \\[12pt]
du & = \frac{-1}{x^2}\,dx
\end{align}
$$
\int\frac{dx}{x^2\sqrt{x^2+1}} = \int \frac{-du}{\sqrt{\frac{1}{u^2}+1}} = \int \frac{-u\,du}{\sqrt{1+u^2}}.
$$
Now write $w=1+u^2$, so $dw=2u\,du$, and thus $-u\,du = dw/2$.
A: Just do it this way: if $u = \frac1{x}$, then $x = \frac1{u}$ and $dx = d(\frac1{u}) = -\frac1{u^2} du$, and then substitute it all into your integral, and you'll get $-\int \frac{udu}{\sqrt{u^2+1}}$ which is easy to solve by making yet another substitution $v = u^2 +1$ or by rewriting it like $-\int \frac{udu}{\sqrt{u^2+1}}$ = $-\int \frac{0.5d(u^2)}{\sqrt{u^2+1}} = -\int \frac{0.5d(u^2+1)}{\sqrt{u^2+1}}$...
A: why dont you go for the substitution of
$ x = tan t $.
then your $ dx = sec^2 t dt $
then your integral would become  
$ \int(1/tan^2t)*(1/sec t )* sec^2t dt $.
i think after that it is an easy one
A: $x=\tan u$  $\Rightarrow$ $dx=(1+\tan^{2}u)du$
So
\begin{align*}
\int\frac{1}{x^{2}\sqrt{x^{2}+1}} & =\int\frac{1}{\tan^{2}u\sqrt{1+\tan^{2}u}}(1+\tan^{2}u)du=\int\frac{\sqrt{1+\tan^{2}u}}{\tan^{2}u}du\\
 & =\int\frac{\sec u}{\tan^{2}u}du
\end{align*}
$$
$$
after some simplification 
\begin{align*}
\int\frac{\sec u}{\tan^{2}u}du & =\int\frac{\cos u}{\sin^{2}u}du
\end{align*}
make substution $s=\sin u\Rightarrow ds=\cos udu$
so 
$$
\begin{align*}
\int\frac{\cos u}{\sin^{2}u}du & =\int\frac{ds}{s^{2}}=\frac{-1}{s}+c
\end{align*}
$$
by back substution $s=\sin u$
$$
\
=\frac{-1}{\sin u}+c
\
$$
and finally by $u=\arctan(x)$
we get
$$
\
=\frac{-1}{\sin (\arctan x)}+c
\
$$
