Giải thích các bước giải:

Câu 86:

Ta có:

$I=\displaystyle\int \dfrac{xdx}{(x+2)^2}$

$\to I=\displaystyle\int \dfrac{x+2-2}{(x+2)^2}dx$

$\to I=\displaystyle\int \dfrac{1}{x+2}-\dfrac{2}{(x+2)^2}dx$

$\to I=\ln|x+2|+\dfrac{2}{x+2}+C$

$\to \displaystyle\int^1_0 \dfrac{xdx}{(x+2)^2}=\ln3-\ln2-\dfrac13$

$\to a=-\dfrac13, b=-1, c=1$

$\to 3a+b+c=-1$

$\to B$

Câu 87:

Ta có:

$I=\displaystyle\int\dfrac{dx}{x\sqrt{1+x^3}}$

$\to I=\displaystyle\int\dfrac{x^2dx}{x^3\sqrt{1+x^3}}$

$\to I=\displaystyle\int\dfrac{d(x^3)}{3x^3\sqrt{1+x^3}}$

Đặt $x^3=t$

$\to I=\displaystyle\int\dfrac{dt}{3t\sqrt{1+t}}$

Đặt $\sqrt{1+t}=u\to 1+t=u^2$

$\to dt=2udu$

$\to I=\displaystyle\int\dfrac{2udu}{3(u^2-1)u}$

$\to I=\displaystyle\int\dfrac{2du}{3(u^2-1)}$

$\to I=\displaystyle\int\dfrac{2du}{3(u-1)(u+1)}$

$\to I=\displaystyle\int\dfrac{(u+1)-(u-1)}{3(u-1)(u+1)}du$

$\to I=\displaystyle\int\dfrac{(u+1)-(u-1)}{3(u-1)(u+1)}du$

$\to I=\displaystyle\int\dfrac13\cdot (\dfrac{1}{u-1}-\dfrac{1}{u+1})du$

$\to I=\dfrac13(\ln|u-1|-\ln|u+1|)+C$

$\to I=-\dfrac{1}{3}\left(\ln \left|\sqrt{1+t}+1\right|-\ln \left|\sqrt{1+t}-1\right|\right)+C$

$\to I=-\dfrac{1}{3}\left(\ln \left|\sqrt{1+x^3}+1\right|-\ln \left|\sqrt{1+x^3}-1\right|\right)+C$

$\to \displaystyle\int^2_1\dfrac{dx}{x\sqrt{1+x^3}}=\dfrac{-\ln \left(2\right)+\ln \left(\sqrt{2}+1\right)-\ln \left(\sqrt{2}-1\right)}{3}$

$\to a=-\dfrac13$