Bài 111823

Question

Tính các tích phân sau :
a)

$M = \int\limits_{e}^{e^3} \left ( \frac{1}{\ln ^2 x} - \frac{1}{\ln x} \right )dx$
b)

$N = \int\limits_{0}^{1} x^2e^{2x}dx.$

sontn87 · Answer 1 · 7/13/2012 11:03:53 AM

a)Đặt

$t=ln x\Rightarrow\int(\frac{1}{ln^2x}-\frac{1}{ln x})dx=\int(\frac{1}{t^2}-\frac{1}{t})de^t$

$=\int(-e^td\frac{1}{t})-\int\frac{e^tdt}{t}=\frac{-e^t}{t}+C$

$\Rightarrow M=\frac{-e^t}{t}|^3_1=e-\frac{e^3}{3}$
b)

$\int x^2e^{2x}dx=\frac{1}{2}\int x^2de^{2x}=\frac{1}{2}x^2e^{2x}-\int xe^{2x}dx$

$=\frac{1}{2}x^2e^{2x}-\frac{1}{2}xde^{2x}=\frac{1}{2}xe^{2x}(x-1)+\frac{1}{2}\int e^{2x}dx$

$=\frac{1}{2}xe^{2x}(x-1)+\frac{1}{4}e^{2x}+C$

$\Rightarrow N=[\frac{1}{2}xe^{2x}(x-1)+\frac{1}{4}e^{2x}]|^1_0=\frac{1}{4}(e^2-1)$

1 Đáp án