Question

Giải hộ giúp tôi bài

3.5 / câu h và g

3.6/ câu c, d,h và k

Mọi người giúp e với ạ !

Thanks

Answer 1

3.5 h)

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

Xét \(\int x\ln (x+1)dx\). Đặt \(\left\{\begin{matrix} u=\ln (x+1)\\ dv=xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dx}{x+1}\\ v=\frac{x^2}{2}\end{matrix}\right.\)

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

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

\(=\frac{x^2\ln (x+1)}{2}-\frac{1}{2}\left(\frac{x^2}{2}-x+\ln |x+1|\right)+c\)

Tương tự, \(\int x\ln (1-x)dx=\frac{x^2\ln (1-x)}{2}-\frac{1}{2}\left (\frac{x^2}{2}+x+\ln |1-x|\right)+c\)

Do đó \(\int x\ln\left (\frac{x+1}{1-x}\right)dx=\frac{x^2\ln \left (\frac{x+1}{1-x}\right)}{2}+x-\frac{1}{2}\ln \left (\frac{x+1}{1-x}\right)+c\)

Answer 2

3.5 g)

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

\(\Rightarrow \int \sqrt{x}\ln ^2xdx=\frac{2\sqrt{x^3}\ln ^2x}{3}-\frac{4}{3}\int \sqrt{x}\ln xdx\)

Xét \(\int \sqrt{x}\ln xdx\)

Đặt \(\left\{\begin{matrix} m=\ln x\\ dn=\sqrt{x}dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} dm=\frac{dx}{x}\\ n=\frac{2\sqrt{x^3}}{3}\end{matrix}\right.\)

\(\Rightarrow \int \sqrt{x}\ln xdx=\frac{2\ln x.\sqrt{x^3}}{3}-\frac{2}{3}\int \sqrt{x}dx\)

\(=\frac{2\ln x.\sqrt{x^3}}{3}-\frac{4\sqrt{x^3}}{9}+c\)

Do đó \(\int \sqrt{x}\ln^2xdx=\frac{2\ln ^2x.\sqrt{x^3}}{3}-\frac{8\ln x.\sqrt{x^3}}{9}+\frac{16\sqrt{x^3}}{27}+c\)

Answer 3

3.6c)

Đặt \(\sqrt{2-5x}=t\Rightarrow x=\frac{2-t^2}{5}\)

Khi đó \(\int x\sqrt{2-5x}dx=\int \frac{t(2-t^2)}{5}d\left (\frac{2-t^2}{5}\right)=\frac{2}{25}\int t^2(t^2-2)dt\)

\(=\frac{2}{25}\left (\frac{t^5}{5}-\frac{2t^3}{3}\right)+c\)

3.6 d)

Đặt \(\left\{\begin{matrix} u=\ln (\sin x)\\ dv=\frac{dx}{\cos ^2x}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{\cos x}{\sin x}dx=\frac{dx}{\tan x}\\ v=\tan x\end{matrix}\right.\)

\(\Rightarrow \int \frac{\ln(\sin x)}{\cos^2x}dx=\tan x\ln (\sin x)-\int dx+c=\tan x\ln (\sin x)-x+c\)

Answer 4

3.6 h)

Đặt \(\sqrt{x}=t\) suy ra \(\int \frac{1}{1-\sqrt{x}}dx=\int \frac{d(t^2)}{1-t}=\int \frac{2tdt}{1-t}\)

\(=2\int \frac{t-1+1}{1-t}dt=-2\int (1+\frac{1}{t-1})dt=-2(t+\ln |t-1|)+c\)

3.6 k)

Ta có \(\int \frac{\sin^3x}{\cos^2x}dx=-\int \frac{\sin^2x}{\cos^2x}d(\cos x)\)

\(=\int \frac{\cos^2x-1}{\cos^2x}d(\cos x)=\int d(\cos x )-\int \frac{d(\cos x)}{\cos ^2x}=\cos x+\frac{1}{\cos x}+c\)