integral of $\arcsin(\sqrt{x/(x+1)})$

Question

I'm trying to solve $$\int \arcsin\bigg( \sqrt{\frac{x}{x + 1}} \bigg) dx$$

I did integration by parts, using $1$ as the term "derivated" and $\arcsin$ as the term "yet to be derivated". I found an expression, made some simplifications, and, to proceed, I need to solve the following integral:

$$\int\frac{x}{(x+1)^2}\cdot \frac{1}{\sqrt{\frac{x}{(x+1)^2}}} dx$$

I thought about "distributing" the sqrt, so, the expression would be

$$\int\frac{x}{(x+1)^2}\cdot \frac{1}{\frac{\sqrt{x}}{\sqrt{(x+1)^2}}} dx$$

$\sqrt{(x+1)^2}$ is equal to $|x + 1|$, but in my list of exercises, the solutions simply equals the expression to $x + 1$. I know how to solve the integral doing that, but shouldn't I separate the integral into two parts?

After all, when you equal the expression to $|x + 1|$ you are assuming $x\geq -1$, which is not always true, since the exercise didn't state that.

I'm confused. I don't know if this is an error or there is something I'm missing.

Thanks in advance

Answer 1

The integral of a function only exists in the domain of said function. For the function $\arcsin(\sqrt{\frac{x}{x+1}})$, the domain is $[0,\infty)$. Thus, the assumption made by the textbook is correct, as in this case, $|x+1| = x+1$ for all $x$ in the domain of this function.

Integrating this function is not hard; you're on the right track. The integral would simplify to $$\int \frac{\sqrt x}{x+1} dx$$ Take $\sqrt x = u \implies \frac{dx}{\sqrt{x}} = 2du$ and the integral becomes $$\int \frac{2u^2}{u^2+1} du$$ which is easily solvable. Can you take it from here?

Answer 2

Let there be a $u(x)$ that we choose later. By chain rule we have $$ (x\arcsin u(x))'=\arcsin u(x)+\frac{x}{\sqrt{1-u^2(x)}}u'(x). $$

Set $u(x)=\sqrt{\frac{x}{1+x}}$ then $u'(x)=\frac{\sqrt{1+x}}{2\sqrt{x}}\frac{1}{(1+x)^2}$ which you insert into the equation above to get $$ (x\arcsin \sqrt{\frac{x}{1+x}})'=\arcsin\sqrt{\frac{x}{1+x}}+\frac{1}{2(1+x)^2} $$ which is $$ \arcsin\sqrt{\frac{x}{1+x}}=(x\arcsin \sqrt{\frac{x}{1+x}})'-\frac{1}{2(1+x)^2}. $$ Use $$ \int\frac{1}{(1+x)^2}dx=\int\frac{1}{1+x}dx-\int\frac{x}{(1+x)}dx=\ln(1+x)-\frac{1}{2}\ln(1+x^2) $$ to obtain $$ \int\arcsin\sqrt{\frac{x}{1+x}}=x\arcsin\sqrt{\frac{x}{1+x}}-\frac{1}{2}\ln(1+x)+\frac{1}{4}\ln(1+x^2)+C. $$

Answer 3

In your integration by parts, a nice trick is instead of integrating $1$ to $x$ you can profitably integrate it to $x+1$ instead, in order to get a simplification with $\arcsin$ derivative.

$\begin{align}\displaystyle\int 1\times\arcsin\left(\sqrt{\frac x{x+1}}\right)\mathop{dx} &=(x+1)\arcsin\left(\sqrt{\frac x{x+1}}\right)-\int \frac{dx}{2\sqrt{x}} \\&=(x+1)\arcsin\left(\sqrt{\frac x{x+1}}\right)-\sqrt{x}+C\end{align}$