Nguyên hàm:
1. \(\int\dfrac{1}{x\ln x}dx\)
Tìm nguyên hàm của hàm số : \(\int\dfrac{x\ln\left(x+\sqrt{x^2+1}\right)}{\sqrt{x^2+1}}dx\)
Lời giải:
Đặt \(u=\ln (x+\sqrt{x^2+1}); dv=\frac{1}{\sqrt{x^2+1}}dx\)
\(\Rightarrow du=\frac{dx}{\sqrt{x^2+1}}; v=\int \frac{x}{\sqrt{x^2+1}}dx=\frac{1}{2}\int \frac{d(x^2+1)}{\sqrt{x^2+1}}=\sqrt{x^2+1}\)
\(\Rightarrow \int \frac{x\ln (x+\sqrt{x^2+1})}{\sqrt{x^2+1}}dx=\int udv=uv-vdu=\sqrt{x^2+1}\ln (x+\sqrt{x^2+1})-\int dx\)
\(=\sqrt{x^2+1}\ln (x+\sqrt{x^2+1})-x+C\)
Tìm các nguyên hàm sau:
a) \(\int (3x^2-2x-4)dx \)
b) \(\int(\sin3x-\cos4x)dx \)
c) \(\int(e^{-3x}-4^x)dx \)
d) \(\int\ln(x)dx \)
e) \(\int(x.e^x)dx \)
f) \(\int(x+1).\sin(x)dx \)
g) \(\int x.\ln(x)dx \)
\(\int\left(3x^2-2x-4\right)dx=x^3-x^2-4x+C\)
\(\int\left(sin3x-cos4x\right)dx=-\dfrac{1}{3}cos3x-\dfrac{1}{4}sin4x+C\)
\(\int\left(e^{-3x}-4^x\right)dx=-\dfrac{1}{3}e^{-3x}-\dfrac{4^x}{ln4}+C\)
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 \(\left\{{}\begin{matrix}u=lnx\\dv=xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=\dfrac{1}{2}x^2\end{matrix}\right.\)
\(\Rightarrow I=\dfrac{1}{2}x^2.lnx-\dfrac{1}{2}\int xdx=\dfrac{1}{2}x^2.lnx-\dfrac{1}{4}x^2+C\)
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) \(\int x\ln\dfrac{1+x}{1-x}dx\)
\(\int(x)ln(x+1)dx\)
Tính nguyên hàm
\(\int xln\left(x+1\right)dx\)
\(\left\{{}\begin{matrix}u=ln\left(x+1\right)\\dv=xdx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=\dfrac{1}{x+1}dx\\v=\dfrac{x^2}{2}\end{matrix}\right.\)
\(\Rightarrow\int xln\left(x+1\right)dx=\dfrac{x^2}{2}.ln\left(x+1\right)-\int\dfrac{x^2}{2}.\dfrac{1}{x+1}dx=\dfrac{x^2}{2}.ln\left(x+1\right)-\dfrac{1}{2}\int\dfrac{x^2}{x+1}dx\)
Xet \(\int\dfrac{x^2}{x+1}dx=\int\dfrac{\left(x+1\right)\left(x-1\right)}{x+1}dx+\int\dfrac{1}{x+1}dx\)
\(=\int\left(x-1\right)dx+\int\dfrac{1}{x+1}dx\)
\(=\dfrac{x^2}{2}-x+ln\left(x+1\right)\)
\(\Rightarrow\int xln\left(x+1\right)dx=\dfrac{x^2}{2}.ln\left(x+1\right)-\dfrac{1}{2}\left(\dfrac{x^2}{2}-x+ln\left(x+1\right)\right)\)
\(\int tan\left(x\right)-ln^{15}\left(cos\left(x\right)\right)dx\)
\(\int\dfrac{x^4+x^2+1}{2x^3+5x^2-7}dx\)
tính nguyên hàm , ai giúp mình 2 bài này với hoặc 1 bài thôi cũng đc ạ , xin cảm ơn nhiều.
Tính nguyên hàm của:
1, \(\int\)\(\dfrac{x^3}{x-2}dx\)
2, \(\int\)\(\dfrac{dx}{x\sqrt{x^2+1}}\)
3, \(\int\)\((\dfrac{5}{x}+\sqrt{x^3})dx\)
4, \(\int\)\(\dfrac{x\sqrt{x}+\sqrt{x}}{x^2}dx\)
5, \(\int\)\(\dfrac{dx}{\sqrt{1-x^2}}\)
a. \(\int\dfrac{x^3}{x-2}dx=\int\left(x^2+2x+4+\dfrac{8}{x-2}\right)dx=\dfrac{1}{3}x^3+x^2+4x+8ln\left|x-2\right|+C\)
b. \(\int\dfrac{dx}{x\sqrt{x^2+1}}=\int\dfrac{xdx}{x^2\sqrt{x^2+1}}\)
Đặt \(\sqrt{x^2+1}=u\Rightarrow x^2=u^2-1\Rightarrow xdx=udu\)
\(I=\int\dfrac{udu}{\left(u^2-1\right)u}=\int\dfrac{du}{u^2-1}=\dfrac{1}{2}\int\left(\dfrac{1}{u-1}-\dfrac{1}{u+1}\right)du=\dfrac{1}{2}ln\left|\dfrac{u-1}{u+1}\right|+C\)
\(=\dfrac{1}{2}ln\left|\dfrac{\sqrt{x^2+1}-1}{\sqrt{x^2+1}+1}\right|+C\)
c. \(\int\left(\dfrac{5}{x}+\sqrt{x^3}\right)dx=\int\left(\dfrac{5}{x}+x^{\dfrac{3}{2}}\right)dx=5ln\left|x\right|+\dfrac{2}{5}\sqrt{x^5}+C\)
d. \(\int\dfrac{x\sqrt{x}+\sqrt{x}}{x^2}dx=\int\left(x^{-\dfrac{1}{2}}+x^{-\dfrac{3}{2}}\right)dx=2\sqrt{x}-\dfrac{1}{2\sqrt{x}}+C\)
e. \(\int\dfrac{dx}{\sqrt{1-x^2}}=arcsin\left(x\right)+C\)
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) \(\int\dfrac{x^3-1}{x+1}dx=\int\dfrac{x^3+1-2}{x+1}dx=\int\left(x^2-x+1-\dfrac{2}{x+1}\right)dx=\dfrac{1}{3}x^3-\dfrac{1}{2}x^2+x-2ln\left|x+1\right|+C\)
\(\Rightarrow A=\dfrac{1}{3};B=\dfrac{1}{2};E=-2\Rightarrow A-B+E=-\dfrac{13}{6}\)
Tìm nguyên hàm của các hàm số sau:
a) \(\int\left(6x-\dfrac{1}{sin^2x}+1\right)dx\)
b) \(\int\dfrac{x^3+2x^2-1}{x^2}dx\)
Tính nguyên hàm I = \(\int\left(x^2+2x\right)ln\left(3x+1\right)dx\)
Tính các nguyên hàm sau :
a) \(\int\left(2x-3\right)\sqrt{x-3}dx\), đặt \(u=\sqrt{x-3}\)
b) \(\int\dfrac{x}{\left(1+x^2\right)^{\dfrac{3}{2}}}dx\) , đặt \(u=\sqrt{x^2+1}\)
c) \(\int\dfrac{e^x}{e^x+e^{-x}}dx\), đặt \(u=e^{2x}+1\)
d) \(\int\dfrac{1}{\sin x-\sin a}dx\)
e) \(\int\sqrt{x}\sin\sqrt{x}dx,\) đặt \(t=\sqrt{x}\)
g) \(\int x\ln\dfrac{x}{1+x}dx\)
a)
Đặt \(u=\sqrt{x-3}\Rightarrow x=u^2+3\)
\(I_1=\int (2x-3)\sqrt{x-3}dx=\int (2u^2+3)ud(u^2+3)=2\int (2u^2+3)u^2du\)
\(\Leftrightarrow I_1=4\int u^4du+6\int u^2du=\frac{4u^5}{5}+2u^3+c\)
b)
\(I_2=\int \frac{xdx}{\sqrt{(x^2+1)^3}}=\frac{1}{2}\int \frac{d(x^2+1)}{\sqrt{(x^2+1)^2}}\)
Đặt \(u=\sqrt{x^2+1}\). Khi đó:
\(I_2=\frac{1}{2}\int \frac{d(u^2)}{u^3}=\int \frac{udu}{u^3}=\int \frac{du}{u^2}=\frac{-1}{u}+c\)
c)
\(I_3=\int \frac{e^xdx}{e^x+e^{-x}}=\int \frac{e^{2x}dx}{e^{2x}+1}=\frac{1}{2}\int\frac{d(e^{2x}+1)}{e^{2x}+1}\)
\(\Leftrightarrow I_3=\frac{1}{3}\ln |e^{2x}+1|+c=\frac{1}{2}\ln|u|+c\)
d)
\(I_4=\int \frac{dx}{\sin x-\sin a}=\int \frac{dx}{2\cos \left ( \frac{x+a}{2} \right )\sin \left ( \frac{x-a}{2} \right )}\)
\(\Leftrightarrow I_4=\frac{1}{\cos a}\int \frac{\cos \left ( \frac{x+a}{2}-\frac{x-a}{2} \right )dx}{2\cos \left ( \frac{x+a}{2} \right )\sin \left ( \frac{x-a}{2} \right )}=\frac{1}{\cos a}\int \frac{\cos \left ( \frac{x-a}{2} \right )dx}{2\sin \left ( \frac{x-a}{2} \right )}+\frac{1}{\cos a}\int \frac{\sin \left ( \frac{x+a}{2} \right )dx}{2\cos \left ( \frac{x+a}{2} \right )}\)
\(\Leftrightarrow I_4=\frac{1}{\cos a}\left ( \ln |\sin \frac{x-a}{2}|-\ln |\cos \frac{x+a}{2}| \right )+c\)
e)
Đặt \(t=\sqrt{x}\Rightarrow x=t^2\)
\(I_5=\int t\sin td(t^2)=2\int t^2\sin tdt\)
Đặt \(\left\{\begin{matrix} u=t^2\\ dv=\sin tdt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=2tdt\\ v=-\cos t\end{matrix}\right.\)
\(\Rightarrow I_5=-2t^2\cos t+4\int t\cos tdt\)
Tiếp tục nguyên hàm từng phần \(\Rightarrow \int t\cos tdt=t\sin t+\cos t+c\)
\(\Rightarrow I_5=-2t^2\cos t+4t\sin t+4\cos t+c\)
g)
Có \(I_6=\int x\ln \left ( \frac{x}{x+1} \right )dx=\int x\ln xdx-\int x\ln (x+1)dx\)
Đặt \(\left\{\begin{matrix} u=\ln x\\ dv=xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dx}{x}\\ v=\frac{x^2}{2}\end{matrix}\right.\Rightarrow \int x\ln xdx=\frac{x^2\ln x}{2}-\int \frac{xdx}{2}\)
\(\Leftrightarrow \int x\ln xdx=\frac{x^2\ln x}{2}-\frac{x^2}{4}+c\)
Tương tự, \(\int x\ln (x+1)dx=\frac{x^2\ln (x+1)}{2}-\int \frac{x^2}{2(x+1)}dx\)
\(=\frac{x^2\ln (x+1)}{2}-\frac{x^2}{4}+\frac{x}{2}-\frac{\ln (x+1)}{2}+c\)
Suy ra \(I_5=\frac{x^2}{2}\ln \frac{x}{x+1}+\frac{1}{2}\ln|x+1|-\frac{x}{2}+c\)