Need a solution to this Integration problem How to evaluate:$\displaystyle\int_{0}^{r}\frac{x^4}{(x^2+y^2)^{\frac{3}{2}}}dx$
I have tried substituting $x =y\tan\ A$, but failed.
 A: Hint. One may assume without loss of generality that $r>0, \, y>0$. Then your change of variable is Ok,
$$
x=y \tan \theta,\quad dx=y (\tan^2 \theta+1)d\theta,
$$ it gives
$$
\int_{0}^{r}\frac{x^4}{(x^2+y^2)^{\frac{3}{2}}}dx=y^2\int_{0}^{\arctan (r/y)}\frac{\sin^4 \theta}{\cos^3 \theta}\:d\theta=y^2\int_{0}^{\arctan (r/y)}\frac{\sin^4 \theta}{(1-\sin^2 \theta)^2}\:\cos \theta\:d\theta
$$  which is easier to evaluate.
A: We have $$\int_{0}^{r}\frac{x^{4}}{\left(x^{2}+y^{2}\right)^{3/2}}dy\stackrel{v=x/y}{=}y^{2}\int_{0}^{r/y}\frac{v^{4}}{\left(v^{2}+1\right)^{3/2}}dv
 $$ $$\overset{v=\tan\left(u\right)}{=}y^{2}\int_{0}^{\arctan\left(r/t\right)}\sin\left(u\right)\tan^{3}\left(u\right)du\tag{1}
 $$ $$=y^{2}\int_{0}^{\arctan\left(r/t\right)}\sec\left(u\right)\tan^{2}\left(u\right)du-y^{2}\int_{0}^{\arctan\left(r/t\right)}\sin\left(u\right)\tan\left(u\right)du\tag{2}
 $$ $$=y^{2}\int_{0}^{\arctan\left(r/t\right)}\sec^{3}\left(u\right)du-y^{2}\int_{0}^{\arctan\left(r/t\right)}\sec\left(u\right)du-y^{2}\int_{0}^{\arctan\left(r/t\right)}\frac{\sin^{2}\left(u\right)}{\cos\left(u\right)}du\tag{3}
 $$ $$=I_{1}-I_{2}-I_{3}
 $$ where $(1)
 $, $(2)$ and $(3)$ follow from the identity $\tan^{2}\left(u\right)+1=\sec^{2}\left(u\right)
 $ and $\tan\left(u\right)=\frac{\sin\left(u\right)}{\cos\left(u\right)}
 $. For $I_{1}
 $ we can use the reduction formula $$\int\sec^{n}\left(x\right)dx=\frac{\sin\left(x\right)\sec^{n-1}\left(x\right)}{n-1}+\frac{n-2}{n-1}\int\sec^{n-2}\left(x\right)dx
 $$ with $n=3
 $. The integral of $\int\sec\left(u\right)du
 $ is easy and the integral $I_{3}
 $ can be integrated observing $$\int\frac{\sin^{2}\left(u\right)}{\cos\left(u\right)}du=\int\frac{1-\cos^{2}\left(u\right)}{\cos\left(u\right)}du
 $$ $$=\int\sec\left(u\right)du-\int\cos\left(u\right)du.
 $$
A: Hint
$$I=\displaystyle\int_{0}^{r}\frac{x^4}{(x^2+y^2)^{\frac{3}{2}}}dx=\displaystyle\int_{0}^{r}\frac{(x^2+y^2)^2}{(x^2+y^2)^{\frac{3}{2}}}dx-2\displaystyle\int_{0}^{r}\frac{x^2y^2}{(x^2+y^2)^{\frac{3}{2}}}dx-y^2\displaystyle\int_{0}^{r}\frac{1}{(x^2+y^2)^{\frac{3}{2}}}dx$$
$$I=\displaystyle\int_{0}^{r}\sqrt{x^2+y^2}dx-2y^2\displaystyle\int_{0}^{r}\frac{x^2}{\sqrt{x^2+y^2}}dx-y^2\displaystyle\int_{0}^{r}\frac{1}{(x^2+y^2)\sqrt{x^2+y^2}}dx$$
$$I=(1-2y^2)\displaystyle\int_{0}^{r}\sqrt{x^2+y^2}dx+2y^4\displaystyle\int_{0}^{r}\frac{1}{\sqrt{x^2+y^2}}dx-y^2\displaystyle\int_{0}^{r}\frac{1}{(x^2+y^2)\sqrt{x^2+y^2}}dx$$
Now set $x=y\tan\theta$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Note that
  $\ds{I \equiv \int_{0}^{r}{x^{4} \over \pars{x^{2} + y^{2}}^{3/2}}\,\dd x =
y^{2}\ \overbrace{\int_{0}^{R}{x^{4} \over \pars{x^{2} + 1}^{3/2}}
\,\dd x}^{\ds{\equiv\ \,\mathcal{I}}}\ \,,\qquad
R \equiv {r \over \verts{y}}}$.


\begin{align}
\,\mathcal{I} & = \int_{0}^{R}{x^{4} \over \pars{x^{2} + 1}^{3/2}}\,\dd x =
\int_{0}^{R}{\bracks{\pars{x^{2} + 1} - 1}^{2}\over \pars{x^{2} + 1}^{3/2}}
\,\dd x
\\[4mm] & =
\int_{0}^{R}\pars{x^{2} + 1}^{1/2}\,\,\dd x -
2\int_{0}^{R}\pars{x^{2} + 1}^{-1/2}\,\dd x + \int_{0}^{R}\pars{x^{2} + 1}^{-3/2}\,\,\dd x
\end{align}

With $\ds{x \equiv \tan\pars{t}}$:
\begin{align}
&\int\bracks{\sec\pars{t} - 2\cos\pars{t} + \cos^{3}\pars{t}}\,\dd t =
\ln\pars{\sec\pars{t} + \tan\pars{t}} -
\int\bracks{1 + \sin^{2}\pars{t}}\cos\pars{t}\,\dd t
\\[4mm] = &\
\ln\pars{\sec\pars{t} + \tan\pars{t}} -
\sin\pars{t} - {1 \over 3}\,\sin^{3}\pars{t}
\\[4mm] & =
\ln\pars{\root{x^{2} + 1} + x} - {x \over \root{x^{2} + 1}} -
{1 \over 3}\,{x^{3} \over \pars{x^{2} + 1}^{3/2}}
\end{align}

\begin{align}
\color{#f00}{\,\mathcal{I}} & =
\color{#f00}{\ln\pars{\root{R^{2} + 1} + R} - {R \over \root{R^{2} + 1}} -
{1 \over 3}\,{R^{3} \over \pars{R^{2} + 1}^{3/2}}}
\end{align}

\begin{align}
\color{#f00}{I} & =
\color{#f00}{y^{2}\ln\pars{\root{r^{2} + y^{2}} + r} -
y^{2}\ln\pars{\verts{y}} -
{y^{2}r \over \root{r^{2} + y^{2}}} -
{1 \over 3}\,{y^{2}r^{3} \over \pars{r^{2} + y^{2}}^{3/2}}}
\end{align}
