Find the $\frac{d}{dx}$ of $g(x)=\int_{2x}^{3x} \frac{u^2-1}{u^2+1}du$ First I split the integral 
$$\int_{2x}^{0} f(u)du+\int_{0}^{3x}f(u)du$$
Then differentiate with respect to x
$$\frac{d}{du}\left[\int_{2x}^{0}f(u)du+ \int_{0}^{3x}f(u)du\right]\frac{dx}{du}$$
In my notes after diff. with respect to x  I  have
$$f'(x)=\frac{4x}{(x^2+1)^2}$$
$$f'(2x)=\frac{8x}{(2x^2+1)^2}$$
$$f'(3x)=\frac{12x}{(3x^2+1)^2}$$
Then for an answer I get 
$$g'(x)=\frac{-2(4x^2-1)}{4x^2+1}+\frac{3(9x^2-1}{9x^2+1}$$
And I can't understand how I arrived at that answer.

Hint
$$\frac{d}{dx}\int_{a(x)}^{b(x)}f(u) \, du=f(b(x)) \frac{d}{dx}(b(x))-f(a(x)) \frac{d}{dx}(a(x))$$
In your case $a(x)=2x, b(x)=3x$ and $f(u)=\frac{u^2-1}{u^2+1}$. You don't need the derivatives of $f$ that you have computed.
