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.
 A: 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? 
A: There's an extremely useful (but apparently rarely taught) tactic for handling integrals involving square-roots of quadratic functions known as Euler substitutions. The appropriate Euler substitution can convert any integral of an algebraic function of the form $R{(x,\sqrt{ax^2+bx+c})}$, where $R$ is a rational function, into just an integral of a pure rational function. 
Applying the technique to your problem, we could use the type-I Euler substitution.
$$\sqrt{x^2+1}=-x+t,$$
$$\implies t=x+\sqrt{x^2+1},~x=\frac{t^2-1}{2t},~$$
$$\implies \mathrm{d}x=\frac{t^2+1}{2t^2}\,\mathrm{d}t.$$
Thus,
$$\begin{align}
\int_{0}^{4}\frac{x^3}{\sqrt{x^2+1}}\,\mathrm{d}x
&=\int_{1}^{4+\sqrt{17}}\frac{(t^2-1)^3}{4t^2(t^2+1)}\cdot\frac{t^2+1}{2t^2}\,\mathrm{d}t\\
&=\frac18\int_{1}^{4+\sqrt{17}}\frac{(t^2-1)^3}{t^4}\,\mathrm{d}t.\\
\end{align}$$
From there, the rest of the problem should be very straightforward, though perhaps algebraically tedious.
A: For this particular integral, you could try trigonometric substitution.
First substitute $x = \tan t,$
then substitute $\sec t = u.$
You should then have
$$\int_0^\sqrt{17} (u^2 - 1) du.$$
(As a check, Wolfram Alpha evaluates this to the exact same value as the original integral.)
A: 
$\qquad$ I then chose to use the substitution $x=\tan y$ and after a series of manipulations I used a second substitution $u=\cos t$ and things got worse. I feel I am doing something wrong.

$x=\tan u~=>~dx=d(\tan u)=\dfrac{du}{\cos^2u}.~$ At the same time, $1+x^2=1+\tan^2u=\dfrac1{\cos^2u}$.
Thus, $~I=\displaystyle\int_0^a\dfrac{\tan^3u}{\bigg(\dfrac1{\cos u}\bigg)}\dfrac{du}{\cos^2u}=\int_0^a\dfrac{\sin^3u}{\cos^4u}du=-\int_0^a\dfrac{1-\cos^2u}{\cos^4u}d(\cos u),~$ which is 
trivial. $~$ Now, $a=\arctan4$, and, after letting $t=\cos u$, the upper limit becomes $\cos(\arctan4)$, 
which, by definition, is the cosine of the angle whose tangent is $4.~$ So, $\tan a=\dfrac{\sin a}{\cos a}=4.~$ But 
$\sin a=\sqrt{1-\cos^2a}.~$ This yields a quadratic equation, whose solution is $\cos a=\dfrac1{\sqrt{17}}$.

Alternately, if you are familiar with hyperbolic functions, let $x=\sinh y.~$ Since $\sinh'y=\cosh y$ 
and $1+\sinh^2y=\cosh^2y$, you will be left with $\displaystyle\int_0^b\sinh^3y~dy=\int_0^b\Big(\cosh^2y-1\Big)~d(\cosh y)$, 
which is also trivial. Here, $b=\sinh^{-1}4$, and, after letting $z=\cosh y$, the new upper limit will be 
$\cosh(\sinh^{-1}4)$, which, by its very definition, is the hyperbolic cosine of the arc whose hyperbolic 
sine is $4.~$ In other words, $\sinh b=4.~$ But $\cosh^2b-\sinh^2b=1\iff\cosh b=\sqrt{1+\sinh^2b}=$
$=\sqrt{1+4^2}=\sqrt{17}$.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
&\color{#66f}{\large\int_{0}^{4}{x^{3} \over \pars{x^{2} + 1}^{1/2}}\,\dd x}
=\int_{x\ =\ 0}^{x\ =\ 4}x^{2}\,\dd\pars{x^{2} + 1}^{1/2}
\\[5mm]&=\left.x^{2}\pars{x^{2} + 1}^{1/2}\,\right\vert_{\, 0}^{\, 4}\ -\
\overbrace{\int_{0}^{4}\pars{x^{2} + 1}^{1/2}\pars{2x}\,\dd x}
^{\ds{\color{#c00000}{x^{2}\ \mapsto\ x}}}\ =\ 
16\root{17} - \int_{0}^{16}\pars{x + 1}^{1/2}\,\dd x
\\[5mm]&=16\root{17} - \left.{2 \over 3}\pars{x + 1}^{3/2}\right\vert_{0}^{16}
=16\root{17} - {2 \over 3}\pars{17^{3/2} - 1}
=\pars{16 - {2 \over 3}\,17}\root{17} + {2 \over 3}
\\[5mm]&=\color{#66f}{\large{14\root{17} + 2 \over 3}}
\approx {\tt 19.9078}
\end{align}
