\(I=\int\limits^e_1xlnxdx+\int\limits^e_1\dfrac{lnx}{x}dx=I_1+I_2\)

Xét \(I_1\) , đặt \(\left\{{}\begin{matrix}u=lnx\\dv=xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=\dfrac{x^2}{2}\end{matrix}\right.\)

\(\Rightarrow I_1=\dfrac{x^2}{2}lnx|^e_1-\int\limits^e_1\dfrac{x}{2}=\dfrac{e^2}{2}-\dfrac{e}{2}+\dfrac{1}{2}\)

Xét \(I_2=\int\limits^e_1\dfrac{lnx}{x}dx=\int\limits^e_1lnx.d\left(lnx\right)=\dfrac{ln^2x}{2}|^e_1=\dfrac{1}{2}\)

\(\Rightarrow I=\dfrac{e^2}{2}-\dfrac{e}{2}+1\)

đặt t=\(t=\sqrt[3]{1+\ln^2x}=>t^3=1+\ln^2x=>3t^2dt=\dfrac{2lnxdx}{x}=>\dfrac{lnxdx}{x}=\dfrac{3t^2}{2}dt=>\int\dfrac{3t^2tdt}{2}=\dfrac{3t^4}{8}+c\)https://www.youtube.com/channel/UCzeAuHrGhk8hUszunoNtayw

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Ta có \(I=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{\ln2.\ln\left(2\tan x\right)}{\sin2x.\ln\left(2\tan x\right)}dx=\ln2\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin2x.\ln\left(2\tan x\right)}+\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin2x}\)

Tính \(\ln2\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin2x.\ln\left(2\tan x\right)}=\frac{\ln2}{2}\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{d\left[\ln\left(2\tan x\right)\right]}{\ln2\left(2\tan x\right)}=\frac{\ln2}{2}\left[\ln\left(\ln\left(2\tan x\right)\right)\right]|^{\frac{\pi}{3}}_{\frac{\pi}{4}}=\frac{\ln2}{2}.\ln\left(\frac{\ln2\sqrt{3}}{\ln2}\right)\)

Tính \(\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin2x}=\frac{1}{2}\ln\left(\tan x\right)|^{\frac{\pi}{3}}_{\frac{\pi}{4}}=\frac{1}{2}\ln\sqrt{3}\)

Vậy \(I=\frac{\ln2}{2}\ln\left(\frac{\ln2\sqrt{3}}{\ln2}\right)+\frac{1}{2}\ln\sqrt{3}\)

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

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

\(\Rightarrow I=A-B=\frac{\ln ^2 2+\ln 2-1}{2}\)

\(I=\int\limits^5_1\left(\frac{x}{\sqrt{x-1}+1}+\frac{\ln x}{\left(x+1\right)^2}\right)dx=\int\limits^5_1\frac{x}{\sqrt{x-1}+1}dx+\int\limits^5_1\frac{\ln x}{\left(x+1\right)^2}dx\)

- Tính \(\int\limits^5_1\frac{x}{\sqrt{x-1}+1}dx\)

Đặt \(t=\sqrt{x-1}\Rightarrow t^2=x-1\Leftrightarrow x=t^2+1\Rightarrow dx=2tdt\)

Đổi cận : Cho x=1 => t=0; x=5=>t=2

\(I_1=\int\limits^2_0\frac{t^2+1}{t+1}.2td=\int\limits^2_0\frac{2t^3+2t}{t+1}dt=\int\limits^2_0\left(2t^2-2t+4-\frac{4}{t+1}\right)dt\)

    \(=\left(\frac{2}{3}t^3-t^2+4t-4\ln\left|x+1\right|\right)|^2_0=\frac{28}{3}-4\ln3\)

\(I_2=\int\limits^5_1\frac{\ln x}{\left(x+1\right)^2}dx\)

Đặt \(\begin{cases}u=\ln x\\dv=\frac{1}{\left(x+1\right)^2}dx\end{cases}\) \(\Rightarrow\begin{cases}du=\frac{1}{x}dx\\v=-\frac{1}{x+1}\end{cases}\)

Ta có \(I_2=-\frac{1}{x+1}\ln x|^5_1+\int\limits^5_1\frac{1}{x\left(x+1\right)}dx=-\frac{1}{6}\ln5+\int\limits^5_1\left(\frac{1}{x}-\frac{1}{x+1}\right)dx\)

\(=-\frac{1}{6}\ln5+\left(\ln\left|x\right|x+1\right)|^5_1=-\frac{1}{6}\ln5+\ln5-\ln6+\ln2=\frac{5}{6}\ln5-\ln3\)

Khi đó \(I=I_1+I_2=\frac{28}{3}+\frac{5}{6}\ln5=5\ln3\)

a)

Ta có \(A=\int ^{\frac{\pi}{4}}_{0}\cos 2x\cos^2xdx=\frac{1}{4}\int ^{\frac{\pi}{4}}_{0}\cos 2x(\cos 2x+1)d(2x)\)

\(\Leftrightarrow A=\frac{1}{4}\int ^{\frac{\pi}{2}}_{0}\cos x(\cos x+1)dx=\frac{1}{4}\int ^{\frac{\pi}{2}}_{0}\cos xdx+\frac{1}{8}\int ^{\frac{\pi}{2}}_{0}(\cos 2x+1)dx\)

\(\Leftrightarrow A=\frac{1}{4}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\sin x+\frac{1}{16}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\sin 2x+\frac{1}{8}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|x=\frac{1}{4}+\frac{\pi}{16}\)

b)

\(B=\int ^{1}_{\frac{1}{2}}\frac{e^x}{e^{2x}-1}dx=\frac{1}{2}\int ^{1}_{\frac{1}{2}}\left ( \frac{1}{e^x-1}-\frac{1}{e^x+1} \right )d(e^x)\)

\(\Leftrightarrow B=\frac{1}{2}\left.\begin{matrix} 1\\ \frac{1}{2}\end{matrix}\right|\left | \frac{e^x-1}{e^x+1} \right |\approx 0.317\)

c)

Có \(C=\int ^{1}_{0}\frac{(x+2)\ln(x+1)}{(x+1)^2}d(x+1)\).

Đặt \(x+1=t\)

\(\Rightarrow C=\int ^{2}_{1}\frac{(t+1)\ln t}{t^2}dt=\int ^{2}_{1}\frac{\ln t}{t}dt+\int ^{2}_{1}\frac{\ln t}{t^2}dt\)

\(=\int ^{2}_{1}\ln td(\ln t)+\int ^{2}_{1}\frac{\ln t}{t^2}dt=\frac{\ln ^22}{2}+\int ^{2}_{1}\frac{\ln t}{t^2}dt\)

Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=\frac{dt}{t^2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=\frac{-1}{t}\end{matrix}\right.\Rightarrow \int ^{2}_{1}\frac{\ln t}{t^2}dt=\left.\begin{matrix} 2\\ 1\end{matrix}\right|-\frac{\ln t+1}{t}=\frac{1}{2}-\frac{\ln 2 }{2}\)

\(\Rightarrow C=\frac{1}{2}-\frac{\ln 2}{2}+\frac{\ln ^22}{2}\)

d)

\(D=\int ^{\frac{\pi}{4}}_{0}\frac{x\sin x+(x+1)\cos x}{x\sin x+\cos x}dx=\int ^{\frac{\pi}{4}}_{0}dx+\int ^{\frac{\pi}{4}}_{0}\frac{x\cos x}{x\sin x+\cos x}dx\)

Ta có:

\(\int ^{\frac{\pi}{4}}_{0}dx=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|x=\frac{\pi}{4}\)

\(\int ^{\frac{\pi}{4}}_{0}\frac{x\cos xdx}{x\sin x+\cos x}=\int ^{\frac{\pi}{4}}_{0}\frac{d(x\sin x+\cos x)}{x\sin x+\cos x}=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|\ln |x\sin x+\cos x|\)

\(=\ln|\frac{\pi\sqrt{2}}{8}+\frac{\sqrt{2}}{2}|\)

Suy ra \(D=\frac{\pi}{4}+\ln|\frac{\pi\sqrt{2}}{8}+\frac{\sqrt{2}}{2}|\)