Question

integral (t+1/t)^3/2 (t^2 -1/t^2)dt

Answer 1

$\displaystyle I=\int \left( \frac {t+1}{t} \right)^{3/2}\cdot\left(\frac{t^2-1}{t^2}\right)\,dt\\ \\ I=\int \left( \frac {t+1}{t} \right)^{3/2}\cdot (1-t^{-2})\,dt\\ \\ I=\int \left( \frac {t+1}{t} \right)^{3/2}\,dt -\int t^{-2}\left( \frac {t+1}{t} \right)^{3/2}\,dt\\ \\ \text{Cambio de variable:}\\ \\ u^2=1+\frac{1}{t}\to dt=-\frac{2u}{(u^2-1)^2}du\\ \\ I=-2\int u^3\cdot \frac{u}{(u^2-1)^2}du-2\int\frac{1}{(u^2-1)^2}\cdot \frac{u}{(u^2-1)^2}du$

$\displaystyle I=-2\int \frac{u^4}{(u^2-1)^2}du-2\int \frac{u}{(u^2-1)^4}du \\ \\ I=-2\int 1+\frac{2u^2+1}{(u^2-1)^2}du-\int \frac{1}{(u^2-1)^4}d(u^2-1) \\ \\ I=-2u-2\int\frac{2u^2+1}{(u^2-1)^2}du+\frac{1}{3(u^2-1)^3}\\ \\ I=-2u+\frac{1}{3(u^2-1)^3}-2\int\frac{3}{4(u - 1)^2} + \frac{1}{4(u - 1)} + \frac{3}{4(u + 1)^2} - \frac{1}{4(u + 1)}du$

$I=-2u+\frac{1}{3(u^2-1)^3} +\frac{3}{2(u-1)}-\frac{1}{2}\ln|u-1|+\frac{3}{2(u+1)}+\frac{1}{2}\ln|u+1|+C$

$\displaystyle I=-2u+\frac{1}{3(u^2-1)^3} +\frac{3u}{u^2-1}+\frac{1}{2}\ln\left|\frac{u+1}{u-1}\right|+C\\ \\ \\ \boxed{I=-2\sqrt{\frac{1+t}{t}}+\frac{t^3}{3}+3t\sqrt{\frac{1+t}{t}}+\ln\left(\sqrt{1+t}+\sqrt{t})+C}$