Calculate the indefinite integral $\int \frac{dx}{({x^2-2x+5})^\frac{3}{2}} $ I have this.
$$\int \frac{dx}{({x^2-2x+5})^\frac{3}{2}} $$
I tried to define -
$ u = x-1 $
$\int \frac{du}{8(\frac{u^2}{4}+1)^{\frac{3}{2}}} = \frac{1}{8}\int \frac{du}{(\frac{u^2}{4}+1)^{\frac{3}{2}}}$
Now, im pretty stuck.
Thanks.
 A: Hint...substitute $x=1+2\tan\theta$
A: You can probably get this far on your own: for any real number $a$,
$$\begin{align}
I{\left(a\right)}
&=\int_{0}^{a}\frac{\mathrm{d}x}{\left(x^2-2x+5\right)^{3/2}}\\
&=\int_{0}^{a}\frac{\mathrm{d}x}{\left[\left(x-1\right)^2+4\right]^{3/2}}\\
&=\int_{-1}^{a-1}\frac{\mathrm{d}y}{\left(y^2+4\right)^{3/2}};~~~\small{\left[x-1=y\right]}\\
&=\int_{-\frac12}^{\frac{a-1}{2}}\frac{2}{\left[4\left(t^2+1\right)\right]^{3/2}}\,\mathrm{d}t;~~~\small{\left[y=2t\right]}\\
&=\frac14\int_{-\frac12}^{\frac{a-1}{2}}\frac{\mathrm{d}t}{\left(t^2+1\right)^{3/2}};~~~\small{\left[y=2t\right]}.\\
\end{align}$$
Next, use the product rule and chain rule to find the following derivative:
$$\begin{align}
\frac{d}{dt}\left[\frac{t}{\sqrt{t^2+1}}\right]
&=\frac{1}{\sqrt{t^2+1}}+t\,\frac{d}{dt}\left[\frac{1}{\sqrt{t^2+1}}\right]\\
&=\frac{1}{\sqrt{t^2+1}}-\frac{t^{2}}{\left(t^2+1\right)^{3/2}}\\
&=\frac{\left(t^2+1\right)}{\left(t^2+1\right)^{3/2}}-\frac{t^{2}}{\left(t^2+1\right)^{3/2}}\\
&=\frac{1}{\left(t^2+1\right)^{3/2}}.\\
\end{align}$$
Thus,
$$\begin{align}
I{\left(a\right)}
&=\frac14\int_{-\frac12}^{\frac{a-1}{2}}\frac{\mathrm{d}t}{\left(t^2+1\right)^{3/2}}\\
&=\frac14\left[\frac{t}{\sqrt{t^2+1}}\right]_{-\frac12}^{\frac{a-1}{2}}\\
&=\frac14\left[\frac{\left(\frac{a-1}{2}\right)}{\sqrt{\left(\frac{a-1}{2}\right)^2+1}}-\frac{\left(-\frac12\right)}{\sqrt{\left(-\frac12\right)^2+1}}\right]\\
&=\frac14\left[\frac{\left(a-1\right)}{\sqrt{\left(a-1\right)^2+2^{2}}}+\frac{1}{\sqrt{1+2^{2}}}\right]\\
&=\frac{\left(a-1\right)}{4\sqrt{\left(a-1\right)^2+4}}+\frac{1}{4\sqrt{5}}.\\
\end{align}$$
Once you've reached the final answer, it's easy to verify that it's correct by checking the derivative
$$\frac{d}{dx}\left[\frac{\left(x-1\right)}{4\sqrt{x^2-2x+5}}\right]=\frac{1}{\left(x^{2}-2x+5\right)^{3/2}}.$$
