Finding a definite integral of an algebraic fraction I need to find the definite integral $\displaystyle\int_0^4 \frac{x^3}{(x^2+1)^{1/2}} dx$.
I first tried to manipulate the fraction algebraically and did not succeed down that path. I then chose to use the substitution $x=\tan y$ and after a series of manipulations I used a second substitution $u=cost$ and things got worse. I feel I am doing something wrong. I need a hint here.

Note that $x^3 = x\cdot x^2$. Now, put $$u = x^2 + 1 \implies du = 2x\,dx \iff \dfrac{du}{2} = x\,dx$$ and $$x^2 = u-1.$$ 
We can't forget the bounds: you can simply change the bounds from $x = 0 \implies u = 1$ to $x = 4 \implies u = 17$. Then there's no need to back substitute.
That gives you $$\begin{align}\int_0^4 \frac{x^3\,dx}{(x^2+1)^{1/2}} & = \frac 12 \int_1^{17} \frac {(u-1)\,du}{u^{1/2}}\\ \\& = \frac 12 \int_1^{17} \left(u^{1/2} - u^{-1/2}\right)\,du\end{align}$$
Can you take it from here? 
