How to compute : $\int \frac{dt}{(t-1)\sqrt{t^2+2t+3}}$?

Question

I have tried to compute this integral , but I couldn't split the x from the square in order to get two simple integrals :

$\int \frac{dt}{(t-1)\sqrt{t^2+2t+3}}$

any advice ?

thanks .

Answer 1

Let $I = \int \frac{1}{(t-1)\sqrt{t^{2}+2t+3}} dt$

Using partial fractions, we get:

$\frac{A}{t-1} +\frac{B}{{\sqrt{t^{2}+2t+3}}} = \frac{1}{(t-1)\sqrt{t^{2}+2t+3}}$

This gives $A= \frac{1}{\sqrt{6}}, B= \frac{\sqrt{2} - 2}{2}$

Then $I = \int \frac{1}{\sqrt{6}(t-1)} + \frac{\sqrt{2} - 2}{2\sqrt{t^{2}+2t+3}} dt$

Can you solve it now?

Edit: In case you get stuck:

$I = \frac{1}{\sqrt{6}}ln(t-1) +\int \frac{\sqrt{2} - 2}{2\sqrt{t^{2}+2t+3}} dt$

Let the $J = \int \frac{\sqrt{2} - 2}{2\sqrt{(t+1)^{2}+2}} dt$

Then use the substitution $\sqrt{2} sinh u = t+1$

This gives $J = \int \frac{\sqrt{2} -2}{2\sqrt{2}}du = \frac{1-\sqrt{2}}{2}u = \frac{1-\sqrt{2}}{2}arsinh(\frac{t+1}{\sqrt{2}})$

Noting that $arsinh x = \frac{1}{\sqrt{1+x^{2}}}$ if you wish to 'simplify' $J$ further, but it's still rather messy so I won't bother.

Then we get $I = \frac{1}{\sqrt{6}}ln(t-1) + \frac{1-\sqrt{2}}{2}arsinh(\frac{t+1}{\sqrt{2}})$