Integrating $\sqrt\frac{x}{{x+a}}$?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 :).

Answer 4

Note that the substitution $x=a\sinh^2t$ is more convenient:

$$\int\sqrt\frac{x}{{x+a}}dx =a\int 2\sinh^2tdt = \frac a2\sinh2t-at+C $$