How to do this without using trigonometric substitution? $$\int \frac {\rho^2}{(\rho^2+ h^2)^\frac 32} d\rho$$
 A: Hint:
Prove that
$$\int\frac1{\sqrt{x^2+1}}\ dx=\ln(x+\sqrt{1+x^2})+c$$
Let $x=tu$ to see that
$$\int\frac1{\sqrt{t^2u^2+1}}\ du=\frac1t\ln(tu+\sqrt{1+t^2u^2})+c$$
Then differentiate with respect to $t$ to see that
$$\int\frac{u^2}{(t^2u^2+1)^{3/2}}\ du=-\frac1t\frac d{dt}\left(\frac1t\ln(tu+\sqrt{1+t^2u^2})+c\right)$$
Be careful, and notice that differentiating $c$ with respect to $t$ is still a constant of integration with respect to $u$.
Now multiply both sides by $t^3$ to finally get
$$\int\frac{u^2}{(u^2+\frac1{t^2})^{3/2}}\ du=-t^2\frac d{dt}\left(\frac1t\ln(tu+\sqrt{1+t^2u^2})\right)+c$$
and then let $t\mapsto\frac1t$ to get your final answer.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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With $\ds{\quad\rho = {h^{2} - t^{2} \over 2t}\,,\quad
t = \root{\rho^{2} + h^{2}} - \rho}$

\begin{align}
&\int{\rho^{2} \over \pars{\rho^{2} + h^{2}}^{3/2}}\,\dd\rho =
\int\bracks{-\,{1 \over t} + {4h^{2}t \over \pars{t^{2} + h^{2}}^{2}}}\,\dd t =
-\ln\pars{t} - {2h^{2} \over t^{2} + h^{2}}
\\[5mm] = &\
-\ln\pars{\root{\rho^{2} + h^{2}} - \rho} -
{2h^{2} \over \pars{\root{\rho^{2} + h^{2}} - \rho}^{2} + h^{2}}
\end{align}
A: I have taken this clue from another answer posted on another question.
First substitute $\rho=\frac1t$, I get:
$$-\int\frac{dt}{t(1+h^2t^2)^{3/2}}$$
Then change $t^2\rightarrow t$, I get:$$\frac{-1}{2}\int\frac{dt}{t(1+th^2)^{3/2}}$$
After this substitute $t=\frac{u^2-1}{h^2}$, which makes the integral:$$-\int\frac{du}{u^2(u^2-1)}$$
After this, apply partial fraction and integrating gives the solution.
A: In addition to the pythagorean identities for circular trig functions, such as
$$ 1 + \tan^2 x = \sec^2 x $$
and for the hyperbolic trig functions, such as
$$ 1 + \sinh^2 x = \cosh^2 x $$
there is a similar identity involving rational functions:
$$ 1 + \left( \frac{y^2 - 1}{2y} \right)^2 = \left( \frac{y^2 + 1}{2y} \right)^2 $$
The various forms of this identity can be computed as needed from
$$ (2y)^2 + (y^2 - 1)^2 = (y^2 + 1)^2 $$
So, for example, when one has
$$ \sqrt{1 + x^2} \, \mathrm{d} x$$
one could make the substitution
$$x = \frac{y^2 - 1}{2y} \qquad \qquad y > 0$$
to get
$$ \left( \frac{y^2 + 1}{2y} \right) \cdot \left( \frac{1}{2} \left( 1 + \frac{1}{y^2} \right) \mathrm{d} y \right) $$
With the square root eliminated, the problem can then be solved via partial fractions. Although in this example the denominator is a power of $y$, so you have the much simpler method of just expanding everything.
Similarly, one could substitute $x = \frac{2y}{y^2 - 1}$ instead.
