Đáp án: $I=2\left(1-2\ln \left(\dfrac{4}{3}\right)\right)$
Giải thích các bước giải:
Ta có:
$I=\displaystyle\int^4_1\dfrac{1}{\sqrt{u}+2}du$
Đặt $\sqrt{u}+2=x\to d(\sqrt{u}+2)=dx, u\Bigg|^4_1\to x\Bigg|^4_3$
$\to \dfrac{1}{2\sqrt{u}}du=dx$
$\to \dfrac{1}{2(x-2)}du=dx$
$\to du=2(x-2)dx$
$\to I=\displaystyle\int^4_3\dfrac{2(x-2)}{x}dx$
$\to I=\displaystyle\int^4_32-\dfrac4xdx$
$\to I=2x-4\ln x\Bigg|^4_3$
$\to I=2\left(1-2\ln \left(\dfrac{4}{3}\right)\right)$
