1, \(\int\dfrac{x}{1-cos2x}dx\)
2, \(\int cos2x.e^{3x}dx\)
3, \(\int\left(2x+1\right)ln^2dx\)
4, \(\int\left(2x-1\right)cosxdx\)
5, \(\int\left(x^2+x+1\right)e^xdx\)
6, \(\int\left(2x+1\right)ln\left(x+2\right)dx\)
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 \(\left\{\begin{matrix} u=\ln t\\ dv=dt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=t\end{matrix}\right.\Rightarrow \int \ln tdt=t\ln t-\int dt=t\ln t-t+c\)
\(\Rightarrow I=(x+1)\ln (x+1)-(x+1)-(x-1)\ln (x-1)+x-1+c\)
\(\Leftrightarrow I=(x+1)\ln(x+1)-(x-1)\ln(x-1)+c\)
Tính các nguyên hàm sau đây :
a) \(\int\left(x+\ln x\right)x^2dx\)
b) \(\int\left(x+\sin^2x\right)\sin xdx\)
c) \(\int\left(x+e^x\right)e^{2x}dx\)
d) \(\int\left(x+\sin x\right)\dfrac{dx}{\cos^2x}\)
e) \(\int\dfrac{e^x\cos x+\left(e^x+1\right)\sin x}{e^x\sin x}dx\)
a) \(\int\left(x+\ln x\right)x^2\text{d}x=\int x^3\text{d}x+\int x^2\ln x\text{dx}\)
\(=\dfrac{x^4}{4}+\int x^2\ln x\text{dx}+C\) (*)
Để tính: \(\int x^2\ln x\text{dx}\) ta sử dụng công thức tính tích phân từng phần như sau:
Đặt \(\left\{{}\begin{matrix}u=\ln x\\v'=x^2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u'=\dfrac{1}{x}\\v=\dfrac{1}{3}x^3\end{matrix}\right.\)
Suy ra:
\(\int x^2\ln x\text{dx}=\dfrac{1}{3}x^3\ln x-\dfrac{1}{3}\int x^2\text{dx}\)
\(=\dfrac{1}{3}x^3\ln x-\dfrac{1}{3}.\dfrac{1}{3}x^3\)
Thay vào (*) ta tính được nguyên hàm của hàm số đã cho bằng:
(*) \(=\dfrac{1}{3}x^3-\dfrac{1}{3}x^3\ln x+\dfrac{1}{9}x^3+C\)
\(=\dfrac{4}{9}x^3-\dfrac{1}{3}x^3\ln x+C\)
b) Đặt \(\left\{{}\begin{matrix}u=x+\sin^2x\\v'=\sin x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}u'=1+2\sin x.\cos x\\v=-\cos x\end{matrix}\right.\)
Ta có:
\(\int\left(x+\sin^2x\right)\sin x\text{dx}=-\left(x+\sin^2x\right)\cos x+\int\left(1+2\sin x\cos^2x\right)\text{dx}\)
\(=-\left(x+\sin^2x\right)\cos x+\int\cos x\text{dx}+2\int\sin x.\cos^2x\text{dx}\)
\(=-\left(x+\sin^2x\right)\cos x+\sin x-2\int\cos^2x.d\left(\cos x\right)\)
\(=-\left(x+\sin^2x\right)\cos x+\sin x-2\dfrac{\cos^3x}{3}+C\)
c) Đặt \(\left\{{}\begin{matrix}u=x+e^x\\v'=e^{2x}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u'=1+e^x\\v=\dfrac{1}{2}e^{2x}\end{matrix}\right.\)
Ta có:
\(\int\left(x+e^x\right)e^{2x}\text{dx}=\dfrac{1}{2}\left(x+e^x\right)e^{2x}-\dfrac{1}{2}\int\left(1+e^x\right)e^{2x}\text{dx}\)
\(=\dfrac{1}{2}\left(x+e^x\right)e^{2x}-\dfrac{1}{2}\int e^{2x}\text{dx}-\dfrac{1}{2}\int e^{3x}\text{dx}\)
\(=\dfrac{1}{2}\left(x+e^x\right)e^{2x}-\dfrac{1}{2}.\dfrac{1}{2}e^{2x}-\dfrac{1}{2}.\dfrac{1}{3}e^{3x}\)
\(=\dfrac{1}{2}xe^{2x}-\dfrac{1}{4}e^{2x}+\dfrac{1}{3}e^{3x}\)
Áp dụng phương pháp tính tích phân, hãy tính các tích phân sau :
a) \(\int\limits^{\dfrac{\pi}{2}}_0x\cos2xdx\)
b) \(\int\limits^{\ln2}_0xe^{-2x}dx\)
c) \(\int\limits^1_0\ln\left(2x+1\right)dx\)
d) \(\int\limits^3_2\left|\ln\left(x-1\right)-\ln\left(x+1\right)\right|dx\)
e) \(\int\limits^2_{\dfrac{1}{2}}\left(1+x-\dfrac{1}{x}\right)e^{x+\dfrac{1}{x}}dx\)
g) \(\int\limits^{\dfrac{\pi}{2}}_0x\cos x\sin^2xdx\)
h) \(\int\limits^1_0\dfrac{xe^x}{\left(1+x\right)^2}dx\)
i) \(\int\limits^e_1\dfrac{1+x\ln x}{x}e^xdx\)
1) \(\int ln^3xdx\)
2) \(\int_0^1\left(x+sin^2x\right)c\text{os}xdx\)
3)\(\int x\left(e^{2x}+\sqrt[3]{x+1}\right)dx\)
Câu 1)
\(I=\int \ln ^3 xdx\). Đặt \(\left\{\begin{matrix} u=\ln ^3x\\ dv=dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{3\ln ^2x}{x}dx\\ v=x\end{matrix}\right.\)
\(\Rightarrow I=x\ln ^3x-3\int \ln^2xdx\)
Tiếp tục nguyên hàm từng phần cho \(\int \ln ^2xdx\) như trên, ta suy ra:
\(\int\ln ^2xdx=x\ln^2x-2\int \ln x dx\).
Tiếp tục nguyên hàm từng phần cho \(\int \ln xdx\Rightarrow \int \ln xdx=x\ln x-x+c\)
Do đó mà \(I=x\ln ^3x-3(x\ln^2x-2x\ln x+2x)+c\)
\(\Leftrightarrow I=x\ln^3x-3x\ln^2x+6x\ln x-6x+c\)
Câu 2)
\(I=\int ^{1}_{0}(x+\sin ^2x)\cos x dx=\int ^{1}_{0}x\cos xdx+\int ^{1}_{0}\sin^2x\cos xdx\)
Đặt \(\left\{\begin{matrix} u=x\\ dv=\cos xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=\sin x\end{matrix}\right.\Rightarrow \int x\cos xdx=x\sin x-\int \sin xdx=x\sin x+\cos x+c\)
\(\Rightarrow \int ^{1}_{0} x\cos xdx=\sin 1+\cos 1-1\)
Còn \(\int ^{1}_{0}\sin^2x\cos xdx=\int ^{1}_{0}\sin ^2xd(\sin x)=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{\sin ^3x}{3}=\frac{\sin^31}{3}\)
\(\Rightarrow I=-1+\sin 1+\cos 1+\frac{\sin ^3 1}{3}\approx 0,0173\)
Câu 3:
Đối với \(\int xe^{2x}dx\)
\(\left\{\begin{matrix} u=x\\ dv=e^{2x}dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=\int e^{2x}dx=\frac{e^{2x}}{2}\end{matrix}\right.\)
\(\Rightarrow \int xe^{2x}=\frac{1}{2}xe^{2x}-\frac{1}{2}\int e^{2x}dx=\frac{1}{2}xe^{2x}-\frac{1}{4}e^{2x}+c\)
Đối với \(\int x\sqrt[3]{x+1}dx=\int \sqrt[3]{(x+1)^4}dx-\int \sqrt{x+1}dx=\frac{3(x+1)^\frac{7}{3}}{7}-\frac{3}{4}(x+1)^{\frac{4}{3}}+c\)
\(\Rightarrow \int x\sqrt[3]{x+1}dx=\frac{3(x+1)^{\frac{4}{3}}(4x-3)}{28}\)
Do đó mà \(\int x(e^{2x}-\sqrt[3]{x+1})dx=\frac{1}{2}xe^{2x}-\frac{1}{4}e^{2x}+\frac{3(x+1)^{\frac{4}{3}}(4x-3)}{28}+c\)
Tìm các nguyên hàm sau :
a) \(I_1=\int\log_2\left(1-3x\right)dx\)
b) \(I_2=\int\left(2x-3\right)\left(\ln x\right)^2dx\)
c)\(I_3=\int\left(4x^2+6x-7\right)\ln xdx\)
d) \(I_4=\int\left(x^2-2x+3\right)a^xdx\) 0<a, \(a\ne1\)
a) Để ý đến công thức đổi cơ số logarit \(\log_2\left(1-3x\right)=\frac{1}{\ln2}\ln\left(1-3x\right)\)
Ta viết nguyên hàm đã cho dưới dạng \(I_1=\frac{1}{\ln2}\int\ln\left(1-3x\right)dx\)
Đặt \(u=\ln\left(1-3x\right)\) , \(dv=dx\)
Khi đó \(du=\frac{-3}{1-3x}dx\), \(v=x\)
Do đó :
\(I_1=\frac{1}{\ln2}\left[x\ln\left(1-3x\right)+3\int\frac{x}{1-3x}dx\right]\)
\(=\frac{1}{\ln2}\left[x\ln\left(1-3x\right)+3\int\frac{1}{3}\left(-1+\frac{1}{1-3x}\right)dx\right]\)
\(=\frac{1}{\ln2}\left[x\ln\left(1-3x\right)-\int dx+\frac{dx}{1-3x}\right]\)
\(=\frac{1}{\ln2}\left[\left(x-\frac{1}{3}\right)\ln\left(1-3x\right)-x\right]+C\)
b) Đặt \(u=\left(\ln x\right)^2\) , \(dv=\left(2x-3\right)dx\)
Khi đó \(du=2\ln x\frac{dx}{x}\) , \(v=x^2-3x\)
Do đó
\(I_2=\left(x^2-3x\right)\left(\ln x\right)^2-2\int\left(x-3\right)\ln xdx\)
\(\int\left(x-3\right)\ln xdx=I_2\)
Ta tính \(I_2\) Ta tìm nguyên hàm bằng cách lấy nguyên hàm từng phàn một làn nữa và thu được.
\(I_2=\left(\frac{1}{2}x^2-3x\right)\ln x-\int\left(\frac{1}{2}x-3\right)dx=\frac{1}{2}\left(x^2-6x\right)\ln x-\frac{1}{4}x^2+3x\)
Từ đó suy ra \(I_2=\left(x^2-3x\right)\left(\ln x\right)^2-\left(x^2-6x\right)\ln x+\frac{1}{2}x^2-6x+C\)
c) Đặt \(u=\ln x\) , \(dv=\left(4x^2+6x-7\right)dx\)
khi đó \(du=\frac{dx}{x}\) , \(v=\int\left(4x^2+6x-7\right)dx=x^4+3x^2-7x\)
Do đó
\(I_3=\left(x^4+3x^2-7x\right)\ln x-\int\frac{x^4+3x^2-7x}{x}dx\)
\(=\left(x^4+3x^2-7x\right)\ln x-\left(\frac{x^4}{4}+\frac{3x^2}{2}-7x\right)+C\)
d) Đặt \(u=x^2-2x+3,a^xdx=dv\). Khi đó \(du=\left(2x-2\right)dx,v=\int a^xdx=\frac{a^x}{\ln a}\)
Do vậy \(I_4=\frac{\left(x^2-2x+3\right)a^x}{\ln a}-\frac{1}{\ln a}\int\left(2x-2\right)a^xdx\)
\(\int\left(2x-2\right)a^xdx=I_4\)*
Ta tính \(I_4\)* bằng cách lấy nguyên hàm từng phần một lần nữa. Ta có :
\(I_4\)*\(=\int\left(2x-2\right)a^xdx=\frac{\left(2x-2\right)a^x}{\ln a}-\frac{1}{\ln a}\int2a^xdx\)
\(=\frac{\left(2x-2\right)a^x}{\ln a}-\frac{2a^x}{\left(\ln a\right)^2}\)
Từ đó suy ra \(I_4=\left[\frac{x^2-2x+3}{\ln a}-\frac{2x-2}{\left(\ln a\right)^2}+\frac{2}{\left(\ln a\right)^3}\right]a^x+C\)
Tính :
a) \(\int\limits^2_{-1}\left(5x^2-x+e^{0,5x}\right)dx\)
b) \(\int\limits^2_{0,5}\left(2\sqrt{x}+\dfrac{3}{x^2}+\cos x\right)dx\)
c) \(\int\limits^2_1\dfrac{dx}{\sqrt{2x+3}}\) (đặt \(t=\sqrt{2x+3}\) )
d) \(\int\limits^2_1\sqrt[3]{3x^3+4}x^2dx\) (đặt \(t=\sqrt[3]{3x^3+4}\) )
e) \(\int\limits^2_{-2}\left(x-2\right)\left|x\right|dx\)
g) \(\int\limits^0_1x\cos xdx\)
h) \(\int\limits^{\dfrac{\pi}{2}}_{\dfrac{\pi}{6}}\dfrac{1+\sin2x+\cos2x}{\sin x+\cos x}dx\)
i) \(\int\limits^{\dfrac{\pi}{2}}_0e^x\sin xdx\)
k) \(\int\limits^e_1x^2\ln^2xdx\)
Áp dụng phương pháp tính nguyên hàm từng phần, hãy tính :
a) \(\int\left(1-2x\right)e^xdx\)
b) \(\int xe^{-x}dx\)
c) \(\int x\ln\left(1-x\right)dx\)
d) \(\int x\sin^2xdx\)
e) \(\int\ln\left(x+\sqrt{1+x^2}\right)dx\)
g) \(\int\sqrt{x}\ln^2xdx\)
h) \(\int x\ln\dfrac{1+x}{1-x}dx\)
Tính các nguyên hàm sau :
a) \(\int x\left(3-x\right)^5dx\)
b) \(\int\left(2^x-3^x\right)^2dx\)
c) \(\int x\sqrt{2-5x}dx\)
d) \(\int\dfrac{\ln\left(\cos x\right)}{\cos^2x}dx\)
e) \(\int\dfrac{x}{\sin^2x}dx\)
\(\int\dfrac{x+1}{\left(x-2\right)\left(x+3\right)}dx\)
h) \(\int\dfrac{1}{1-\sqrt{x}}dx\)
i) \(\int\sin3x\cos2xdx\)
k) \(\int\dfrac{\sin^3x}{\cos^2x}dx\)
l) \(\int\dfrac{\sin x\cos x}{\sqrt{a^2\sin^2x+b^2\cos^2x}}dx\) (\(a^2\ne b^2\))
Tính các nguyên hàm.
a)\(\int\dfrac{2dx}{x^2-5x}=A\ln\left|x\right|+B\ln\left|x-5\right|+C\) . Tìm 2A-3B.
b)\(\int\dfrac{x^3-1}{x+1}\)dx=\(Ax^3-Bx^2+x+E\ln\left|x+1\right|+C\).Tính A-B+E
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) \(\int\dfrac{x^3-1}{x+1}dx=\int\dfrac{x^3+1-2}{x+1}dx=\int\left(x^2-x+1-\dfrac{2}{x+1}\right)dx=\dfrac{1}{3}x^3-\dfrac{1}{2}x^2+x-2ln\left|x+1\right|+C\)
\(\Rightarrow A=\dfrac{1}{3};B=\dfrac{1}{2};E=-2\Rightarrow A-B+E=-\dfrac{13}{6}\)