Câu 1)

Ta có \(I=\int ^{1}_{0}\frac{dx}{\sqrt{3+2x-x^2}}=\int ^{1}_{0}\frac{dx}{4-(x-1)^2}\).

Đặt \(x-1=2\cos t\Rightarrow \sqrt{4-(x-1)^2}=\sqrt{4-4\cos^2t}=2|\sin t|\)

Khi đó:

\(I=\int ^{\frac{2\pi}{3}}_{\frac{\pi}{2}}\frac{d(2\cos t+1)}{2\sin t}=\int ^{\frac{2\pi}{3}}_{\frac{\pi}{2}}\frac{2\sin tdt}{2\sin t}=\int ^{\frac{2\pi}{3}}_{\frac{\pi}{2}}dt=\left.\begin{matrix} \frac{2\pi}{3}\\ \frac{\pi}{2}\end{matrix}\right|t=\frac{\pi}{6}\)

Câu 3)

\(K=\int ^{3}_{2}\ln (x^3-3x+2)dx=\int ^{3}_{2}\ln [(x+2)(x-1)^2]dx\)

\(=\int ^{3}_{2}\ln (x+2)d(x+2)+2\int ^{3}_{2}\ln (x-1)d(x-1)\)

Xét \(\int \ln tdt\): Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=dt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=t\end{matrix}\right.\Rightarrow \int \ln t dt=t\ln t-t\)

\(\Rightarrow K=\left.\begin{matrix} 3\\ 2\end{matrix}\right|(x+2)[\ln (x+2)-1]+2\left.\begin{matrix} 3\\ 2\end{matrix}\right|(x-1)[\ln (x-1)-1]\)

\(=5\ln 5-4\ln 4-1+4\ln 2-2=5\ln 5-4\ln 2-3\)

Bài 2)

\(J=\int ^{1}_{0}x\ln (2x+1)dx\). Đặt \(\left\{\begin{matrix} u=\ln (2x+1)\\ dv=xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{2dx}{2x+1}\\ v=\frac{x^2}{2}\end{matrix}\right.\)

Khi đó:

\(J=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{x^2\ln (2x+1)}{2}-\int ^{1}_{0}\frac{x^2}{2x+1}dx\)\(=\frac{\ln 3}{2}-\frac{1}{4}\int ^{1}_{0}(2x-1+\frac{1}{2x+1})dx\)

\(=\frac{\ln 3}{2}-\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{x^2-x}{4}-\frac{1}{8}\int ^{1}_{0}\frac{d(2x+1)}{2x+1}=\frac{\ln 3}{2}-\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{\ln (2x+1)}{8}\)

\(=\frac{\ln 3}{2}-\frac{\ln 3}{8}=\frac{3\ln 3}{8}\)

Câu 5)

\(J=\underbrace{\int ^{3}_{1}\frac{3dx}{(x+1)^2}}_{A}+\underbrace{\int ^{3}_{1}\frac{\ln xdx}{(x+1)^2}}_{B}\)

Ta có: \(A=\int ^{3}_{1}\frac{3d(x+1)}{(x+1)^2}=\left.\begin{matrix} 3\\ 1\end{matrix}\right|\frac{-3}{x+1}=\frac{3}{4}\)

\(B=\int ^{3}_{1}\frac{\ln xdx}{(x+1)^2}=\left.\begin{matrix} 3\\ 1\end{matrix}\right|\frac{-\ln x}{x+1}+\int ^{3}_{1}\frac{dx}{x(x+1)}=\frac{-\ln 3}{4}+\left.\begin{matrix} 3\\ 1\end{matrix}\right|(\ln |x|-\ln|x+1|)\)

\(B=\frac{-\ln 3}{4}+(\ln 3-\ln 4)+\ln 2=\frac{3}{4}\ln 3-\ln 2\)

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

1/ \(\int\limits^e_1\left(x+\frac{1}{x}+\frac{1}{x^2}\right)dx=\left(\frac{x^2}{2}+lnx-\frac{1}{x}\right)|^e_1=\frac{e^2}{2}-\frac{1}{e}+\frac{3}{2}\)

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

\(=\left(\frac{2}{5}.x^{\frac{5}{2}}+x\right)|^2_1=\frac{8\sqrt{2}-7}{5}\)

3/

\(\int\limits^2_1\frac{2x^3-4x+5}{x}dx=\int\limits^2_1\left(2x^2-4+\frac{5}{x}\right)dx=\left(\frac{2}{3}x^3-4x+5lnx\right)|^2_1=\frac{2}{3}+5ln2\)

4/ \(\int\limits^2_1x^2\left(3x-1\right)\frac{2}{x}dx=\int\limits^2_1\left(6x^2-2x\right)dx=\left(2x^3-x^2\right)|^2_1=11\)

Lời giải:

Đặt \(x=2\sin t( \frac{-\pi}{2}\leq t\leq \frac{\pi}{2})\)

Khi đó \(A=\int^{\frac{3}{2}}_{0}\frac{dx}{\sqrt{(4-x^2)^9}}=\int ^{\sin ^-1\left(\frac{3}{4}\right)}_{0}\frac{d(2\sin t)}{\sqrt{(4-4\sin^2 x)^9}}=\frac{1}{2^8}\int ^{\sin ^-1\left(\frac{3}{4}\right)}_{0}\frac{dt}{\cos^8 x}\)

Xét \(\int ^{\sin ^-1\left(\frac{3}{4}\right)}_{0}\frac{dt}{\cos^8 x}=\int ^{\sin ^-1\left(\frac{3}{4}\right)}_{0}\frac{d(\tan x)}{\cos ^6x}=\int ^{\sin ^-1\left(\frac{3}{4}\right)}_{0}\frac{(\sin^2x+\cos^2x)^3d(\tan x)}{\cos^6 x}\)

\(=\int ^{\sin ^-1\left(\frac{3}{4}\right)}_{0}(\tan^2 x+1)^3d(\tan x)=\int ^{\sin ^-1\left(\frac{3}{4}\right)}_{0}(\tan^6x+1+3\tan ^4x+3\tan ^2x)d(\tan x)\)

\(=\left.\begin{matrix} \sin^{-1}\left(\frac{3}{4}\right)\\ 0\end{matrix}\right|\left ( \frac{\tan ^7x}{7}+\tan x+\frac{3\tan^5x}{5}+\tan^3x \right )\)

\(\Rightarrow A=\left.\begin{matrix} \sin^{-1}\left(\frac{3}{4}\right)\\ 0\end{matrix}\right|\left ( \frac{\tan ^7x}{7}+\tan x+\frac{3\tan^5x}{5}+\tan^3x \right ).\frac{1}{2^8}\approx 0,015862\)

P/s: Kiểm tra kết quả tại http://www.wolframalpha.com/input/?i=integral+of+%5Csqrt%7B(4-x%5E2)%5E%7B-9%7D%7D+from+0+to+%5Cfrac%7B3%7D%7B2%7D

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

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 \(I=\frac{-1}{2\sqrt{3}}v+c\)

Câu 6)

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

Câu 8)

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

\(\Leftrightarrow I=\int \ln (x+1)d(x+1)-\int \ln (x-1)d(x-1)\)

Xét \(\int \ln tdt\) ta có:

Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=dt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=t\end{matrix}\right.\Rightarrow \int \ln tdt=t\ln t-\int dt=t\ln t-t+c\)

\(\Rightarrow I=(x+1)\ln (x+1)-(x+1)-(x-1)\ln (x-1)+x-1+c\)

\(\Leftrightarrow I=(x+1)\ln(x+1)-(x-1)\ln(x-1)+c\)

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

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