Tính tích phân $I = \displaystyle\int_1^{3} x \ln x\,dx$.
A
$I = \dfrac{9}{2}\ln 3 - 2$
✓
B
$I = \dfrac{9}{2}\ln 3 + 2$
C
$I = \dfrac{9}{2}\ln 3 - 4$
D
$I = \dfrac{9}{2}\ln 3$
LỜI GIẢI
Đặt $u = \ln x$, $dv = x\,dx$ $\Rightarrow du = \dfrac{1}{x}dx$, $v = \dfrac{x^2}{2}$.
$I = \left.\dfrac{x^2}{2}\ln x\right|_1^{3} - \displaystyle\int_1^{3} \dfrac{x^2}{2} \cdot \dfrac{1}{x}\,dx = \dfrac{9}{2}\ln 3 - \displaystyle\int_1^{3} \dfrac{x}{2}\,dx$.
$\displaystyle\int_1^{3} \dfrac{x}{2}\,dx = \left.\dfrac{x^2}{4}\right|_1^{3} = \dfrac{9 - 1}{4} = 2$.
Vậy $I = \dfrac{9}{2}\ln 3 - 2$.
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