How do i find this : $\int \frac{1}{(x+a) \sqrt{x+b}}\ dx$, where $a > b > 0$?

Question

Is there someone show me how do I find : $$\int \frac{1}{(x+a) \sqrt{x+b}}\ dx$$, where $$a > b > 0$$ ?

I tried to make it as sum of fraction to be easier but sorry i didn't up

Thank you for any help .

Answer 1

Use the substitution $u^2 = x + b$.

Let

$I = \int\frac{1}{(x+ a)\sqrt{x + b}}\,dx $ Let \begin{align*} &u^2 = x + b\implies x + a = u^2 + a - b\\ &2u\,du = dx \end{align*} Then \begin{align*} I &= 2\int\frac{1}{u^2 + a - b}\,du\\ \\ &= \frac{2}{\sqrt{a - b}}\tan^{-1}\left(\frac{u}{\sqrt{a - b}}\right) + C\\ \\ &= \frac{2}{\sqrt{a - b}}\tan^{-1}\left(\frac{\sqrt{x + b}}{\sqrt{a - b}}\right) + C \end{align*}

Answer 2

Let $u=\sqrt{x+b}$ and go from there.

Answer 3

If we let $u=\sqrt{x+b}$ then $x = u^2 - b$ and $dx = 2u du$.

Thus the substitution yields:

$$\int \frac{2u du}{u(u^2 + (a-b))} = \int \frac{2}{u^2 + (a-b)} du$$

Now since $a > b$, this means $u^2 + (a-b)$ is irreducible. Thus you should use a formula for this case, or compute the antiderivative directly.