lm giú mk nhá

Question

$\int\limits_{2}^{5}\frac{ln(\sqrt{x-1}+1)}{x-1+\sqrt{x-1}}$ dx

score 4 · Answer 1

đặt

$t=\sqrt{x-1}\Rightarrow 2tdt=dx$

đổi cận

$x=2\Rightarrow t=1, x=5\Rightarrow t=2$

I=2 $\int\limits_{1}^{2}\frac{ln(t+1)}{t^2+t}tdt=2\int\limits_{1}^{2}\frac{ln(t+1)}{(t+1)}dt=2\int\limits_{1}^{2}ln(t+1) d[ln(t+1)]$
đặt u=ln(t+1) $\Rightarrow du=d[ln(t+1)]$
$t=1\Rightarrow u=ln2,t=2\Rightarrow u=ln3$
$\Rightarrow I=2\int\limits_{ln2}^{ln3}udu=u^2\left| {\begin{matrix} ln3\\ ln2 \end{matrix}} \right.=ln^23-ln^22$