Integration of 3x/... function - Substitution How do I integrate the function $\frac{3x}{\sqrt(16x^4+8x^2+1)}$ ?
I thought about doing this:
$\int \frac{3x}{\sqrt(16x^4+8x^2+1)} dx = \int \frac{3x}{(2x)^2+(1)^2}$
then: Substitute $2x = u \rightarrow dx = du/2  $
However, I do not come closer to the end, as you can see:
$\int \frac{3x}{\sqrt(16x^4+8x^2+1)} dx = \int \frac{3u}{2*2*(u^2+1)}du = \int \frac{3u}{(2u)^2 + (2)^2} $

Let$$u=x^2$$
then,
$$du=2xdx$$
Hence
$$I=\int \frac{1.5}{\sqrt(16u^2+8u+1)} du $$
or
$$I=\int \frac{1.5}{4u+1} du $$
Use this now:
$$\int \frac{dx}{ax+b} =  {1\over a}{{ln|ax+b|}}+C$$
