1) \(\int ln\frac{\left(1+s\text{inx}\right)^{1+c\text{os}x}}{1+c\text{os}x}dx\)
2) \(\int\left(xlnx\right)^2dx\)
3) \(\int\frac{3xcosx+2}{1+cot^2x}dx\)
4)\(\int\frac{2}{c\text{os}2x-7}dx\)
5)\(\int\frac{1+x\left(2lnx-1\right)}{x\left(x+1\right)^2}dx\)
6) \(\int\frac{1-x^2}{\left(1+x^2\right)^2}dx\)
7)\(\int e^x\frac{1+s\text{inx}}{1+c\text{os}x}dx\)
8) \(\int ln\left(\frac{x+1}{x-1}\right)dx\)
9)\(\int\frac{xln\left(1+x\right)}{\left(1+x^2\right)^2}dx\)
10) \(\int\frac{ln\left(x-1\right)}{\left(x-1\right)^4}dx\)
11)\(\int\frac{x^3lnx}{\sqrt{x^2+1}}dx\)
12)\(\int\frac{xe^x}{_{ }\left(e^x+1\right)^2}dx\)
13) \(\int\frac{xln\left(x+\sqrt{1+x^2}\right)}{x+\sqrt{1+x^2}}dx\)
giúp mk đc con nào thì giúp nha
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 \(\left\{\begin{matrix} u=\ln t\\ dv=dt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=t\end{matrix}\right.\Rightarrow \int \ln tdt=t\ln t-\int dt=t\ln t-t+c\)
\(\Rightarrow I=(x+1)\ln (x+1)-(x+1)-(x-1)\ln (x-1)+x-1+c\)
\(\Leftrightarrow I=(x+1)\ln(x+1)-(x-1)\ln(x-1)+c\)
Câu 5)
\(I=\int \frac{1+x(2\ln x-1)}{x(x+1)^2}dx=\int \frac{dx}{x(x+1)^2}-\int \frac{dx}{(x+1)^2}+\int \frac{2\ln xdx}{(x+1)^2}\)
\(=\int \left ( \frac{1}{x}-\frac{1}{x+1}-\frac{1}{(x+1)^2} \right )dx+\int \frac{2\ln xdx}{(x+1)^2}-\int \frac{dx}{(x+1)^2}\)
\(\Leftrightarrow I=\ln|x|-\ln|x+1|+\frac{2}{x+1}+\int \frac{2\ln x dx}{(x+1)^2}+c\)
Đối với \(\int \frac{2\ln x}{(x+1)^2}dx\), ta đặt \(\left\{\begin{matrix} u=\ln x\\ dv=\frac{dx}{(x+1)^2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dx}{x}\\ v=\frac{-1}{x+1}\end{matrix}\right.\)
\( \Rightarrow \int \frac{2\ln xdx}{(x+1)^2}=\frac{-2\ln x}{x+1}+\int \frac{1}{x(x+1)}dx=\frac{-2\ln x}{x+1}+2\ln |x|-2\ln|x+1|+c\)
\(\Rightarrow I=3\ln |x|-3\ln|x+1|+\frac{2}{x+1}-\frac{2\ln x}{x+1}+c\)
Lời giải
Đặt \(x=2t\). Biến đổi ta có:
\(I=\int e^{2t}\frac{(\sin t+\cos t)^2}{\cos^2t}dt=\int e^{2t}(\tan t+1)^2dt\)
\(\Leftrightarrow I=\int e^{2t}\tan^2t dt+2\int e^{2t}\tan tdt+\int e^{2t}dt\)
Đặt \(\left\{\begin{matrix} u=e^{2t}\\ dv=\tan^2tdt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=2e^{2t}dt\\ v=\int \tan^2tdt=\int \frac{1-\cos^2t}{\cos^2t}dt=\tan t-t\end{matrix}\right.\)
\(\Rightarrow I=e^{2t}(\tan t-t)-2\int (\tan t-t)e^{2t}dt+2\int e^{2t}\tan tdt+\int e^{2t}dt\)
\(\Leftrightarrow I=e^{2t}(\tan t-t)+2\int e^{2t}tdt+\int e^{2t}dt\)
Đặt \(\left\{\begin{matrix} k=t\\ dl=e^{2t}dt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} dk=dt\\ l=\int e^{2t}dt=\frac{e^{2t}}{2}\end{matrix}\right.\)
\(\Rightarrow I=e^{2t}(\tan t-t)+te^{2t}-\int e^{2t}dt+\int e^{2t}dt\)
Hay \(I=e^{2t}\tan t+c=e^x\tan \frac{x}{2}+c\)
Câu 11)
\(I=\int \frac{x^3\ln x}{\sqrt{x^2+1}}=\int \frac{x^2\ln (x^2)d(x^2)}{4\sqrt{x^2+1}}\). Đặt \(\sqrt{x^2+1}=t\rightarrow x^2=t^2-1\).
Khi đó \(I=\int \frac{(t^2-1)\ln(t^2-1)d(t^2-1)}{4t}=\frac{1}{2}\int (t^2-1)\ln (t^2-1)dt\)
Đặt \(\left\{\begin{matrix} u=\ln (t^2-1)\\ dv=(t^2-1)dt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{2t}{t^2-1}dt\\ v=\frac{t^3}{3}-t\end{matrix}\right.\)
\(\Rightarrow 2I=\ln (t^2-1)\left(\frac{t^3}{3}-t\right)-2\int \left(\frac{t^3}{3}-t\right)\frac{t}{t^2-1}dt\)
Đối với \(\int \frac{t^4}{t^2-1}dt\)
\(\int \frac{t^4}{t^2-1}dt=\int (t^2+1+\frac{1}{t^2-1})dt=\frac{t^3}{3}+t+\frac{\ln|t-1|-\ln|t+1|}{2}+c\)
Đối với \(\int \frac{t^2}{t^2-1}dt\)
\(\int \frac{t^2}{t^2-1}dt=\int (1+\frac{1}{t^2-1})dt=t+\frac{\ln|t-1|-\ln|t+1|}{2}+c\)
Do đó mà \(I=\frac{1}{2}\left[\left(\frac{t^3}{3}-t\right)\ln (t^2-1)-\frac{2t^3}{9}+\frac{4t}{3}+\frac{2(\ln|t-1|-\ln|t+1|)}{3}\right]+c\)
Câu 10.
Đặt \(x-1=t\Rightarrow I=\int \frac{\ln t}{t^4}dt\)
Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=\frac{dt}{t^4}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=\frac{-1}{3t^3}\end{matrix}\right.\Rightarrow I=\frac{-\ln t}{3t^3}+\int \frac{dt}{3t^4}=\frac{-\ln t}{3t^3}-\frac{1}{9t^3}+c\)
Câu 12)
Đặt \(\left\{\begin{matrix} u=x\\ dv=\frac{e^x}{(e^x+1)^2}dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=\int \frac{e^x}{(e^x+1)^2}dx=\int\frac{d(e^x+1)}{(e^x+1)^2}=-\frac{1}{e^x+1}\end{matrix}\right.\)
\(\Rightarrow I=\frac{-x}{e^x+1}+\int \frac{dx}{e^x+1}=\frac{-x}{e^x+1}+\int \frac{d(e^x)}{e^x(e^x+1)}=\frac{-x}{e^x+1}+\ln|e^x|-\ln|e^x+1|+c\)
Hay \(I=\frac{xe^x}{e^x+1}-\ln|e^x+1|+c\)
Câu 3)
\(I=\int \frac{3x\cos x+2}{1+\cot ^2x}dx=\int (3x\cos x+2)\sin^2xdx=3\int x\cos x\sin^2xdx+2\int sin^2xdx \)
Ta có:
\(2\int \sin^2xdx=\int (1-\cos 2x)dx=x-\frac{\sin 2x}{2}+c\)
Đối với \(3\int x\cos x\sin^2xdx\): Đặt \(\left\{\begin{matrix} u=x\\ dv=\cos x\sin^2xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=\int \sin^2xd(\sin x)=\frac{\sin^3x}{3}\end{matrix}\right.\)
\(\Rightarrow 3\int x\cos x\sin^2xdx=x\sin^3x-\int \sin^3xdx=x\sin^3x-\int \frac{3\sin x-\sin 3x}{4}dx\)
\(=x\sin^3x+\frac{3}{4}\cos x-\frac{\cos 3x}{12}+c\)
Akai ơi, bà cho tôi giải ké 1 câu với :)))
1)
\(ln\frac{\left(1+sinx\right)^{1+cosx}}{1+cosx}\\ =\left(1+cosx\right)ln\left(1+sinx\right)-ln\left(1+cosx\right)\\ =\left[ln\left(1+sinx\right)-ln\left(1+cosx\right)\right]+cosx\cdot ln\left(1+sinx\right)\\ =ln\frac{1+sinx}{1+cosx}+cosx\cdot ln\left(1+sinx\right)\)
Ta có: \(I=\int ln\frac{1+sinx}{1+cosx}dx+\int cosx\cdot ln\left(1+sinx\right)dx\)
+) \(B=\int cosx\cdot ln\left(1+sinx\right)dx\\ =\int ln\left(1+sinx\right)d\left(1+sinx\right)=\int ln\left(t\right)dt=...\)
+) \(A=\int ln\frac{1+sinx}{1+cosx}dx\)
Đặt: \(f\left(x\right)=\frac{1+sinx}{1+cosx}\Rightarrow f\left(t\right)=\frac{1+sint}{1+cost}\)
Hoàn toàn có thể đặt: \(f\left(x\right)=e^{f\left(t\right)}\)
Đạo hàm 2 vế ta có: \(f'\left(x\right)dx=e^{f\left(t\right)}.e^t.dt\)
Ta có: \(A=\int ln\left(e^{f\left(t\right)}\right).e^t.dt=\int e^t.f\left(t\right)dt=\int e^t.\frac{1+sint}{1+cost}dt\)
(đến đây thì làm giống câu 7 thôi)
