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\(\int\left(3x^2-2x-4\right)dx=x^3-x^2-4x+C\)
\(\int\left(sin3x-cos4x\right)dx=-\dfrac{1}{3}cos3x-\dfrac{1}{4}sin4x+C\)
\(\int\left(e^{-3x}-4^x\right)dx=-\dfrac{1}{3}e^{-3x}-\dfrac{4^x}{ln4}+C\)
d. \(I=\int lnxdx\)
Đặt \(\left\{{}\begin{matrix}u=lnx\\dv=dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=x\end{matrix}\right.\)
\(\Rightarrow u=x.lnx-\int dx=x.lnx-x+C\)
e. Đặt \(\left\{{}\begin{matrix}u=x\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=dx\\v=e^x\end{matrix}\right.\)
\(\Rightarrow I=x.e^x-\int e^xdx=x.e^x-e^x+C\)
f.
Đặt \(\left\{{}\begin{matrix}u=x+1\\dv=sinxdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=dx\\v=-cosx\end{matrix}\right.\)
\(\Rightarrow I=-\left(x+1\right)cosx+\int cosxdx=-\left(x+1\right)cosx+sinx+C\)
g.
Đặt \(\left\{{}\begin{matrix}u=lnx\\dv=xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=\dfrac{1}{2}x^2\end{matrix}\right.\)
\(\Rightarrow I=\dfrac{1}{2}x^2.lnx-\dfrac{1}{2}\int xdx=\dfrac{1}{2}x^2.lnx-\dfrac{1}{4}x^2+C\)
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 \(I=\frac{-1}{2\sqrt{3}}v+c\)
a) \(\int\dfrac{2dx}{x^2-5x}=\int\left(\dfrac{-2}{5x}+\dfrac{2}{5\left(x-5\right)}\right)dx=-\dfrac{2}{5}ln\left|x\right|+\dfrac{2}{5}ln\left|x-5\right|+C\)
\(\Rightarrow A=-\dfrac{2}{5};B=\dfrac{2}{5}\Rightarrow2A-3B=-2\)
b) \(\int\dfrac{x^3-1}{x+1}dx=\int\dfrac{x^3+1-2}{x+1}dx=\int\left(x^2-x+1-\dfrac{2}{x+1}\right)dx=\dfrac{1}{3}x^3-\dfrac{1}{2}x^2+x-2ln\left|x+1\right|+C\)
\(\Rightarrow A=\dfrac{1}{3};B=\dfrac{1}{2};E=-2\Rightarrow A-B+E=-\dfrac{13}{6}\)
Biến đổi: ʃ\(\int\dfrac{1dx}{cosx\dfrac{\sqrt{2}}{2}\left(cosx-sinx\right)}=\int\dfrac{\sqrt{2}dx}{cos^2x\left(1-tanx\right)}=\int\dfrac{\sqrt{2}d\left(tanx\right)}{1-tanx}=-\sqrt{2}\ln trituyetdoi\left(1-tanx\right)\)
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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 \(\left\{\begin{matrix} u=\ln t\\ dv=\frac{dt}{t^2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=\frac{-1}{t}\end{matrix}\right.\Rightarrow \int ^{2}_{1}\frac{\ln t}{t^2}dt=\left.\begin{matrix} 2\\ 1\end{matrix}\right|-\frac{\ln t+1}{t}=\frac{1}{2}-\frac{\ln 2 }{2}\)
\(\Rightarrow C=\frac{1}{2}-\frac{\ln 2}{2}+\frac{\ln ^22}{2}\)
Câu a)
\(\int \frac{1}{\cos^4x}dx=\int \frac{\sin ^2x+\cos^2x}{\cos^4x}dx=\int \frac{\sin ^2x}{\cos^4x}dx+\int \frac{1}{\cos^2x}dx\)
Xét \(\int \frac{1}{\cos^2x}dx=\int d(\tan x)=\tan x+c\)
Xét \(\int \frac{\sin ^2x}{\cos^4x}dx=\int \frac{\tan ^2x}{\cos^2x}dx=\int \tan^2xd(\tan x)=\frac{\tan ^3x}{3}+c\)
Vậy :
\(\int \frac{1}{\cos ^4x}dx=\frac{\tan ^3x}{3}+\tan x+c\)
\(\Rightarrow \int ^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{dx}{\cos^4 x}=\)\(\left.\begin{matrix} \frac{\pi}{3}\\ \frac{\pi}{6}\end{matrix}\right|\left ( \frac{\tan ^3 x}{3}+\tan x+c \right )=\frac{44}{9\sqrt{3}}\)
Câu b)
\(\int \frac{(x+1)^2}{x^2+1}dx=\int \frac{x^2+1+2x}{x^2+1}dx=\int dx+\int \frac{2xdx}{x^2+1}\)
\(=x+c+\int \frac{d(x^2+1)}{x^2+1}=x+\ln (x^2+1)+c\)
Do đó:
\(\int ^{1}_{0}\frac{(x+1)^2}{x^2+1}dx=\left.\begin{matrix} 1\\ 0\end{matrix}\right|(x+\ln (x^2+1)+c)=\ln 2+1\)
Câu c)
\(\int \frac{x^2+2\ln x}{x}dx=\int xdx+2\int \frac{2\ln x}{x}dx\)
\(=\frac{x^2}{2}+c+2\int \ln xd(\ln x)\)
\(=\frac{x^2}{2}+c+\ln ^2x\)
\(\Rightarrow \int ^{2}_{1}\frac{x^2+2\ln x}{x}dx=\left.\begin{matrix} 2\\ 1\end{matrix}\right|\left ( \frac{x^2}{2}+\ln ^2x +c \right )=\frac{3}{2}+\ln ^22\)
Câu d)
\(\int^{2}_{1} \frac{x^2+3x+1}{x^2+x}dx=\int ^{2}_{1}dx+\int ^{2}_{1}\frac{2x+1}{x^2+x}dx\)
\(=\left.\begin{matrix} 2\\ 1\end{matrix}\right|x+\int ^{2}_{1}\frac{d(x^2+x)}{x^2+x}=1+\left.\begin{matrix} 2\\ 1\end{matrix}\right|\ln |x^2+x|=1+\ln 6-\ln 2\)
\(=1+\ln 3\)
