Bài 111620

Question

Tính các tích phân sau:
$ A = \int\limits \left ( \frac{\ln x}{x} \right )^2 dx$
$ B = \int\limits x^2 \log _2 (x-1)dx$

score 0 · Answer 1

a.$\left\{ \begin{array}{l} u=ln^2x\\ dv=\frac{dx}{x^2} \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} du=\frac2xlnxdx\\ v=-\frac1x \end{array} \right.$
$A=-\frac1xln^2x+2\int\limits\frac{lnx}{x^2}dx= -\frac1xln^2x+ 2I$
$\left\{ \begin{array}{l} u=lnx\\ dv=\frac{dx}{x^2} \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} du=\frac{dx}x\\ v=-\frac1x \end{array} \right.$
$I=-\frac1xlnx+\int\limits\frac{dx}{x^2}=-\frac1xlnx-\frac1x+C$
$\Rightarrow A= -\frac{\ln ^2 x + \ln x^2 + 2}{x} + C$
b.$\left\{ \begin{array}{l} u=log_2(x)(x-1)\\ dv=x^2dx \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} du=\frac1{(x-1)ln2}\\ v=\frac{x^3}3 \end{array} \right.$
$B= \frac{1}{3} x^3 \log _2 ( x-1) -\int\limits\frac{x^3}{(x-1)ln8}dx$
$= \frac{1}{3} x^3 \log _2 ( x-1) -\frac1{ln8}\int\limits\left ( x^2+x+1+\frac1{x-1} \right )dx $
$ = \frac{1}{3} x^3 \log _2 ( x-1) - \frac{1}{\ln 8 }\left ( \frac{1}{3}x^3 + \frac{1}{2} x^2 + x + \ln | x-1| \right ) + C$

Bài 111620

1 Đáp án

Lý thuyết liên quan

Bài 111620

1 Đáp án

Liên quan

Tích phân từng phần

Lý thuyết liên quan