Question

Answer 1

Đặt \(\sqrt{3x+1}=u\Rightarrow x=\dfrac{u^2-1}{3}\)

\(dx=\dfrac{2u}{3}du\)

\(\left\{{}\begin{matrix}x=0\Rightarrow u=1\\x=1\Rightarrow u=2\end{matrix}\right.\)

\(I=\int\limits^2_1\dfrac{u}{\dfrac{u^2-1}{3}-5}.\dfrac{2u}{3}du=\int\limits^2_1\dfrac{2u^2}{u^2-16}du\)

\(=2\int\limits^2_1du+32\int\limits^2_1\dfrac{1}{u^2-16}du=2\int\limits^2_1du+4\int\limits^2_1\left(\dfrac{1}{u-4}-\dfrac{1}{u+4}\right)du\)

\(=2u|^2_1+4.ln\left|\dfrac{u-4}{u+4}\right||^2_1=2+4.ln\left(\dfrac{5}{9}\right)\)