Câu hỏi của Huỳnh Như

Question

Tính các tích phân:

a) $\int\limits^1_0$$\dfrac{xe^x+1+x}{e^x+1}$dx

b)$\int\limits^{\dfrac{\pi}{2}}_0$$\dfrac{1-\sin\left(x\right)}{1+\cos\left(x\right)}$dx

c)$\int\limits^2_1$\(\dfrac{\l...

Akai Haruma · Accepted Answer

Câu a)

$I=\int ^{1}_{0}\frac{x(e^x+1)+1}{e^x+1}dx=\int ^{1}_{0}xdx+\int ^{1}_{0}\frac{dx}{e^x+1}$

$=\left.\begin{matrix}
1\

0\end{matrix}\right|\frac{x^2}{2}+\int ^{1}_{0}\frac{d(e^x)}{e^x(e^x+1)}=\frac{1}{2}+\left.\begin{matrix}
1\

0\end{matrix}\right|\ln\left | \frac{e^x}{e^x+1} \right |$

$\Leftrightarrow I=\frac{3}{2}+\ln 2-\ln (e+1)$

Câu d)

$I=\int ^{e}_{1}\ln(x+1)d(x)=\int ^{e}_{1}\ln (x+1)d(x+1)$

Đặt $\left\{\begin{matrix}
u=\ln (x+1)\

dv=d(x+1)\end{matrix}\right.\Rightarrow \left\{\begin{matrix}
du=\frac{d(x+1)}{x+1}\

v=x+1\end{matrix}\right.$

$\Rightarrow I=\left.\begin{matrix}
e\

1\end{matrix}\right|(x+1)\ln (x+1)-\int ^{e}_{1}d(x+1)$

$=(e+1)\ln \left ( \frac{e+1}{e} \right )-2\ln \left (\frac{2}{e}\right )$Câu trả lời: Câu a)

$I=\int ^{1}_{0}\frac{x(e^x+1)+1}{e^x+1}dx=\int ^{1}_{0}xdx+\int ^{1}_{0}\frac{dx}{e^x+1}$

$=\left.\begin{matrix}
1\

0\end{matrix}\right|\frac{x^2}{2}+\int ^{1}_{0}\frac{d(e^x)}{e^x(e^x+1)}=\frac{1}{2}+\left.\begin{matrix}
1\

0\end{matrix}\right|\ln\left | \frac{e^x}{e^x+1} \right |$

$\Leftrightarrow I=\frac{3}{2}+\ln 2-\ln (e+1)$

Câu d)

$I=\int ^{e}_{1}\ln(x+1)d(x)=\int ^{e}_{1}\ln (x+1)d(x+1)$

Đặt $\left\{\begin{matrix}
u=\ln (x+1)\

dv=d(x+1)\end{matrix}\right.\Rightarrow \left\{\begin{matrix}
du=\frac{d(x+1)}{x+1}\

v=x+1\end{matrix}\right.$

$\Rightarrow I=\left.\begin{matrix}
e\

1\end{matrix}\right|(x+1)\ln (x+1)-\int ^{e}_{1}d(x+1)$

$=(e+1)\ln \left ( \frac{e+1}{e} \right )-2\ln \left (\frac{2}{e}\right )$

Akai Haruma · Answer

Câu b)

Đặt $\tan \frac{x}{2}=t$. Ta có:

$\left\{\begin{matrix}
dt=d\left ( \tan \frac{x}{2} \right )=\frac{1}{2\cos ^2\frac{x}{2}}dx=\frac{t^2+1}{2}dx\rightarrow dx=\frac{2dt}{t^2+1}\\

\cos x=\frac{1-t^2}{t^2+1}\end{matrix}\right.$

$ I=\underbrace{\int ^{\frac{\pi}{2}}_{0}\frac{1}{1+\cos x}dx}_{A}+\underbrace{\int ^{\frac{\pi}{2}}_{0}\frac{d(\cos x)}{\cos x+1}}_{B}$

Có $B=\int ^{\frac{\pi}{2}}_{0}\frac{d(\cos x+1)}{\cos x+1}=\left.\begin{matrix}
\frac{\pi}{2}\

0\end{matrix}\right|\ln |\cos x+1|=-\ln 2$

$A=\int ^{1}_{0}\frac{2dt}{(t^2+1)\frac{2}{t^2+1}}=\int ^{1}_{0}dt=1$

$\Rightarrow I=A+B=1-\ln 2$Câu trả lời: Câu b)

Đặt $\tan \frac{x}{2}=t$. Ta có:

$\left\{\begin{matrix}
dt=d\left ( \tan \frac{x}{2} \right )=\frac{1}{2\cos ^2\frac{x}{2}}dx=\frac{t^2+1}{2}dx\rightarrow dx=\frac{2dt}{t^2+1}\\

\cos x=\frac{1-t^2}{t^2+1}\end{matrix}\right.$

$ I=\underbrace{\int ^{\frac{\pi}{2}}_{0}\frac{1}{1+\cos x}dx}_{A}+\underbrace{\int ^{\frac{\pi}{2}}_{0}\frac{d(\cos x)}{\cos x+1}}_{B}$

Có $B=\int ^{\frac{\pi}{2}}_{0}\frac{d(\cos x+1)}{\cos x+1}=\left.\begin{matrix}
\frac{\pi}{2}\

0\end{matrix}\right|\ln |\cos x+1|=-\ln 2$

$A=\int ^{1}_{0}\frac{2dt}{(t^2+1)\frac{2}{t^2+1}}=\int ^{1}_{0}dt=1$

$\Rightarrow I=A+B=1-\ln 2$

Akai Haruma · Answer

Câu c)

Xét $I=\underbrace{\int ^{2}_{1}\frac{\ln xdx}{x}}_{A}-\underbrace{\int ^{2}_{1}\frac{\ln xdx}{x^2}}_{B}$

Có $ A=\int ^{2}_{1}\ln xd(\ln x)=\left.\begin{matrix}
2\

1\end{matrix}\right|\frac{\ln ^2 x}{2}=\frac{\ln ^2 2}{2}$

Với $B$ đặt $\left\{\begin{matrix}
u=\ln x\

dv=\frac{dx}{x^2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix}
du=\frac{dx}{x}\

v=\frac{-1}{x}\end{matrix}\right.$

$\Rightarrow B=\left.\begin{matrix}
2\

1\end{matrix}\right|\frac{-\ln x}{x}+\int ^{2}_{1}\frac{dx}{x^2}=\left.\begin{matrix}
2\

1\end{matrix}\right|\left ( \frac{-\ln x}{x}-\frac{1}{x} \right )=\frac{1}{2}-\frac{\ln 2}{2}$

$\Rightarrow I=A-B=\frac{\ln ^2 2+\ln 2-1}{2}$Câu trả lời: Câu c)