Integrating $\sqrt\frac{x}{{x+a}}$? How can I integrate this expression?
$$\sqrt\frac{x}{{x+a}}$$
I substituted $x=a\tan^2t$. After substituting, I reach
$$\int \sec x\tan^2x$$
Is this way correct?
Thanks in advance.
 A: EDIT: Natasha J, here is your new answer!
Let $$t=\dfrac{x}{x+a}, \ \dfrac{at}{(1-t)}=x$$
$$\int \sqrt{\dfrac{x}{x+a}}dx=a\int \sqrt{t}\dfrac{1(1-t)-(-1)t}{(1-t)^2}dt$$
$$=a\int\dfrac{\sqrt{t}}{(1-t)^2}dt$$
by uv-substitution
$$=\dfrac{a\sqrt{t}}{1-t}-\dfrac{a}{2}\int \dfrac{1}{(1-t)\sqrt{t}}dt$$ where the left side integral is, with $t=u^2$ $$-\dfrac{a}{2}\int \dfrac{1}{1-u}+\dfrac{1}{1+u}du=-\dfrac{a}{2}\left(\ln|1+u|-\ln|1-u|\right)+C$$
which gives
$$\boxed{\int\sqrt{\dfrac{x}{x+a}}dx=\sqrt{x+a}\sqrt{x}-\dfrac{a}{2}\ln\dfrac{|\sqrt{x+a}+\sqrt{x}|}{|\sqrt{x+a}-\sqrt{x}|}+C}$$ Also,
$$0=\lim_{a\to 0}\dfrac{a}{2}\ln\dfrac{|\sqrt{x+a}+\sqrt{x}|}{|\sqrt{x+a}-\sqrt{x}|}$$
It was simpler than it seemed. Using similar steps,
$$\int \sqrt[n]{\dfrac{ax+b}{cx+d}}dx$$ can be known.
A: One method is to just substitute
$$t=\sqrt{\frac{x}{x+a}}$$
that is $$ x = \frac{at^2}{t^2-1}$$
$$ dx = \frac{-2at}{(t^2-1)^2} dt $$
We get
$$ \int \sqrt{\frac{x}{x+a}} dx = \int \frac{-2at^2}{(t^2-1)^2} dt$$
This is an integral of a rational function and can be solved using standard methods.
Another method is to use substitution given by the condition
$$ \sqrt{x^2+ax} = u-x$$
which can be solved to get
$$ x = \frac{u^2}{2u+a}$$
$$ \sqrt{x^2+ax} = \frac{u(u+a)}{2u+a}$$
$$ dx = \frac{2u(u+a)}{(2u+a)^2}dt $$
We get then
$$ \int \sqrt{\frac{x}{x+a}} dx  = \int \frac{x}{\sqrt{x^2+ax}} dx = \int \frac{2u^2}{(2u+a)^2}du $$
which also can be solved using standard methods.
These substitutions are special cases of so-called Euler subsitutions.
A: I will answer to your exact question. YES!, it is correct. The substitution transform your integral to $$2a\displaystyle\int \sec x\tan^2x\, dx.$$ Then you can simplify it further using the fact that $\sec x\tan^2x=\sec^3x-\sec x.$ Powers of secant is quit different from other integrals if that format, but can integrate using a reduction formula. First let $$I_n=\displaystyle\int \sec^nx\, dx,$$ then $I_0=x,\quad I_1=\ln|\sec x+\tan x|$ and $I_2=\tan x$ are the simplest known cases up to constants. For $n\ge 2$ use integration by parts with $u=\sec^{n-2}x$ and $dv=\sec^2x dx$ to get $$I_n=\sec^{n-2}x\tan x-(n-2)\displaystyle\int \sec^{n-2}x\tan^2 x\, dx.$$ Since $\tan^2 x=\sec^2x-1,$ we get a simplified recursive formula $$I_n=\dfrac{\sec^{n-2}x\tan x}{n-1}-\left(\dfrac{n-2}{n-1}\right)I_{n-2}.$$ Hope you can get your solution from here :).
A: Note that the substitution $x=a\sinh^2t$ is more convenient:
$$\int\sqrt\frac{x}{{x+a}}dx
=a\int 2\sinh^2tdt
= \frac a2\sinh2t-at+C
$$
