Đặt \(u=\left(x^3-2x^x+3x+1\right)\Rightarrow du=\left(3x^2-4x+3\right)dx;dv=\frac{dx}{e^{2x}}\Rightarrow v=-\frac{2}{e^{2x}}\)

Ta được : \(-\frac{2}{e^{2x}}\left(x^3-2x^2+3x+1\right)|^1_0+2\int\limits^1_0\left(\frac{3x^2-4x+3}{e^{2x}}\right)dx=2-\frac{6}{e^2}+2J\)

Tương tự ta tính J

Đăth \(u_1=\left(3x^2-4x+3\right)\Rightarrow du_1=\left(6x-4\right)dx;dv_1=\frac{dx}{e^{2x}}\Rightarrow v_1=-\frac{2}{e^{2x}}\left(1\right)\)

Do đó :

\(J=-\frac{2}{e^{2x}}\left(3x^2-4x+3\right)|^1_0+2\int\limits^1_0\frac{6x-4}{e^{2x}}dx=6-\frac{4}{e^2}+2K\left(2\right)\)

Ta tính K :

\(K=\int\limits^1_0\frac{6x-4}{e^{2x}}dx\)

Đặt \(u_2=6x-4\Rightarrow du_2=6dx;dv_2=\frac{dx}{e^{2x}}\Rightarrow v_2=-\frac{2}{e^{2x}}\)

Do đó : \(K=-\frac{2}{e^{2x}}\left(x-4\right)|^1_0+2\int\limits^1_0\frac{6dx}{e^{2x}}=\frac{6}{e^x}-8-6\frac{1}{e^{2x}}|^1_0\left(\frac{1}{e^2}-1\right)=-2\left(3\right)\)

Thay (3) vào (2) 

\(J=6-\frac{4}{e^2}+2\left(-2\right)=2-\frac{4}{e^2}\)

Lại thay vào (1) ta có :

\(I=2-\frac{6}{e^2}+2\left(2-\frac{4}{e^2}\right)=6-\frac{14}{e^2}\)

1/ \(I=\int\limits^1_0\dfrac{2x+1}{x^2+x+1}dx=\int\limits^1_0\dfrac{d\left(x^2+x+1\right)}{x^2+x+1}=ln\left|x^2+x+1\right||^1_0=ln3\)

2/ \(\int\limits^{\dfrac{1}{2}}_0\dfrac{5x}{\left(1-x^2\right)^3}dx=-\dfrac{5}{2}\int\limits^{\dfrac{1}{2}}_0\dfrac{d\left(1-x^2\right)}{\left(1-x^2\right)^3}=\dfrac{5}{4}\dfrac{1}{\left(1-x^2\right)^2}|^{\dfrac{1}{2}}_0=\dfrac{35}{36}\)

3/ \(\int\limits^1_0\dfrac{2x}{\left(x+1\right)^3}dx\Rightarrow\) đặt \(x+1=t\Rightarrow x=t-1\Rightarrow dx=dt;\left\{{}\begin{matrix}x=0\Rightarrow t=1\\x=1\Rightarrow t=2\end{matrix}\right.\)

\(I=\int\limits^2_1\dfrac{2\left(t-1\right)dt}{t^3}=\int\limits^2_1\left(\dfrac{2}{t^2}-\dfrac{2}{t^3}\right)dt=\left(\dfrac{-2}{t}+\dfrac{1}{t^2}\right)|^2_1=\dfrac{1}{4}\)

4/ \(\int\limits^1_0\dfrac{4x-2}{\left(x^2+1\right)\left(x+2\right)}dx\)

Kĩ thuật chung là tách và sử dụng hệ số bất định như sau:

\(\dfrac{4x-2}{\left(x^2+1\right)\left(x+2\right)}=\dfrac{ax+b}{x^2+1}+\dfrac{c}{x+2}=\dfrac{\left(a+c\right)x^2+\left(2a+b\right)x+2b+c}{\left(x^2+1\right)\left(x+2\right)}\)

\(\Rightarrow\left\{{}\begin{matrix}a+c=0\\2a+b=4\\2b+c=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b=0\\a=-c=2\end{matrix}\right.\)

\(\Rightarrow I=\int\limits^1_0\left(\dfrac{2x}{x^2+1}-\dfrac{2}{x+2}\right)dx=\int\limits^1_0\dfrac{d\left(x^2+1\right)}{x^2+1}-2\int\limits^1_0\dfrac{d\left(x+2\right)}{x+2}=ln\dfrac{8}{9}\)

5/ \(\int\limits^1_0\dfrac{x^2dx}{x^6-9}\Rightarrow\) đặt \(x^3=t\Rightarrow3x^2dx=dt\Rightarrow x^2dx=\dfrac{1}{3}dt;\left\{{}\begin{matrix}x=0\Rightarrow t=0\\x=1\Rightarrow t=1\end{matrix}\right.\)

\(I=\dfrac{1}{3}\int\limits^1_0\dfrac{dt}{t^2-9}=\dfrac{1}{18}\int\limits^1_0\left(\dfrac{1}{t-3}-\dfrac{1}{t+3}\right)dt=\dfrac{1}{18}ln\left|\dfrac{t-3}{t+3}\right||^1_0=-\dfrac{1}{18}ln2\)

6/ Tương tự câu 4, sử dụng hệ số bất định ta tách được:

\(\int\limits^2_1\dfrac{2x-1}{x^2\left(x+1\right)}dx=\int\limits^2_1\left(\dfrac{3x-1}{x^2}-\dfrac{3}{x+1}\right)dx=\int\limits^2_1\left(\dfrac{3}{x}-\dfrac{1}{x^2}-\dfrac{3}{x+1}\right)dx\)

\(=\left(3ln\left|\dfrac{x}{x+1}\right|+\dfrac{1}{x}\right)|^2_1=3ln\dfrac{4}{3}-\dfrac{1}{2}\)

\(\int\limits^1_0\dfrac{dx}{\left(x^2+3x+2\right)^2}=\int\limits^1_0\left(\dfrac{1}{x+1}-\dfrac{1}{x+1}\right)^2dx\)

\(=\int\limits^1_0\dfrac{dx}{\left(x+1\right)^2}+\int\limits^1_0\dfrac{dx}{\left(x+2\right)^2}-2\int\limits^1_0\dfrac{dx}{\left(x+1\right)\left(x+2\right)}\)

\(=-\dfrac{1}{x+1}\left|^1_0-\dfrac{1}{x+2}\right|^1_0-2\int\limits^1_0\dfrac{dx}{x+1}+2\int\limits^1_0\dfrac{dx}{x+2}\)

\(=\dfrac{2}{3}-4ln2+2ln3\)

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

\(I=\int\limits^1_0\left(x+e^{2x}\right)xdx=\int\limits^1_0x^2dx+\int\limits^1_0xe^{2x}dx=I_1+I_2\)

\(I_1=\int\limits^1_0x^2dx=\frac{x^3}{3}|^1_0=\frac{1}{3}\)

Đặt \(\begin{cases}dv=e^{2x}dx\\u=x\end{cases}\) ta có \(\begin{cases}v=\frac{e^{2x}}{2}\\du=dx\end{cases}\)

\(I_2=\frac{xe^{2x}}{2}|^1_0-\int\limits^1_0\frac{e^{2x}}{2}dx=\left(\frac{xe^{2x}}{2}-\frac{e^{2x}}{4}\right)|^1_0=\frac{e^2+1}{4}\)

\(I=I_1+I_2=\frac{e^2+1}{4}+\frac{1}{3}=\frac{3e^2+7}{12}\)

\(I=\int\limits^1_0\frac{x+1-1dx}{\left(x+1\right)^3}=\int\limits^1_0\frac{dx}{\left(x+1\right)^2}-\int\limits^1_0\frac{dx}{\left(x+1\right)^3}=x+1|^1_0+\frac{1}{2\left(x+1\right)^2}|^1_0=\frac{1}{8}\)