Question

[Phan thu trang]

1) \(\int ln\frac{\left(1+s\text{inx}\right)^{1+c\text{os}x}}{1+c\text{os}x}dx\)

2) \(\int\left(xlnx\right)^2dx\)

3) \(\int\frac{3xcosx+2}{1+cot^2x}dx\)

4)\(\int\frac{2}{c\text{os}2x-7}dx\)

5)\(\int\frac{1+x\left(2lnx-1\right)}{x\left(x+1\right)^2}dx\)

6) \(\int\frac{1-x^2}{\left(1+x^2\right)^2}dx\)

7)\(\int e^x\frac{1+s\text{inx}}{1+c\text{os}x}dx\)

8) \(\int ln\left(\frac{x+1}{x-1}\right)dx\)

9)\(\int\frac{xln\left(1+x\right)}{\left(1+x^2\right)^2}dx\)

10) \(\int\frac{ln\left(x-1\right)}{\left(x-1\right)^4}dx\)

11)\(\int\frac{x^3lnx}{\sqrt{x^2+1}}dx\)

12)\(\int\frac{xe^x}{_{ }\left(e^x+1\right)^2}dx\)

13) \(\int\frac{xln\left(x+\sqrt{1+x^2}\right)}{x+\sqrt{1+x^2}}dx\)

giúp mk đc con nào thì giúp nha

Answer 1

[Akai Haruma]

Câu 2)

Đặt \(\left\{\begin{matrix} u=\ln ^2x\\ dv=x^2dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=2\frac{\ln x}{x}dx\\ v=\frac{x^3}{3}\end{matrix}\right.\Rightarrow I=\frac{x^3}{3}\ln ^2x-\frac{2}{3}\int x^2\ln xdx\)

Đặt \(\left\{\begin{matrix} k=\ln x\\ dt=x^2dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} dk=\frac{dx}{x}\\ t=\frac{x^3}{3}\end{matrix}\right.\Rightarrow \int x^2\ln xdx=\frac{x^3\ln x}{3}-\int \frac{x^2}{3}dx=\frac{x^3\ln x}{3}-\frac{x^3}{9}+c\)

Do đó \(I=\frac{x^3\ln^2x}{3}-\frac{2}{9}x^3\ln x+\frac{2}{27}x^3+c\)

Answer 2

[Akai Haruma]

Câu 3:

\(I=\int\frac{2}{\cos 2x-7}dx=-\int\frac{2}{2\sin^2x+6}dx=-\int\frac{dx}{\sin^2x+3}\)

Đặt \(t=\tan\frac{x}{2}\Rightarrow \left\{\begin{matrix} \sin x=\frac{2t}{t^2+1}\\ dx=\frac{2dt}{t^2+1}\end{matrix}\right.\)

\(\Rightarrow I=-\int \frac{2dt}{(t^2+1)\left ( \frac{4t^2}{(t^2+1)^2}+3 \right )}=-\int\frac{2(t^2+1)dt}{3t^4+10t^2+3}=-\int \frac{2d\left ( t-\frac{1}{t} \right )}{3\left ( t-\frac{1}{t} \right )^2+16}=\int\frac{2dk}{3k^2+16}\)

Đặt \(k=\frac{4}{\sqrt{3}}\tan v\). Đến đây dễ dàng suy ra \(I=\frac{-1}{2\sqrt{3}}v+c\)