Integrating $\frac{1}{(x^2+b)^{3/2}}$?

Question

How to integrate $$\int\frac{1}{(x^2+b)^{3/2}}dx$$ using the hyperbolic sine substitution ?

Answer 1

Hint: Take $x = \sqrt b \sinh t $ then $dx = \sqrt b \cosh t\, dt$ and

$$\int \frac{\sqrt b \cosh t }{(b \cosh^2 t)^{3/2}} \, dt = \int \frac{1}{b \cosh^2 t} \, dt = \color {#f05}{\frac{\tanh t}{b} + C} $$

Answer 2

Hint: Put $x=\sqrt b \sinh t$ for $b>0$ so the integral becomes $$ \int \frac{1}{b^{3/2}\cosh^3 t}\sqrt b \cosh t \,\mathrm{d}t=.... $$

Answer 3

hint: if $b = 0$, it is a trivial case... if $b > 0$, let $u = \sqrt{b}\cdot \tan x$. if $b < 0$, let $u = \sqrt{b} \cdot \sec x$. Can you continue?