giupf mik voi ạ
giupf mik voi ạ
a. \(\int cosx.sin^3xdx=\int sin^3x.d\left(sinx\right)=\dfrac{sin^4x}{4}+C\)
b. \(\int sinx.cos^5xdx=-\int cos^5x.d\left(cosx\right)=-\dfrac{cos^6x}{6}+C\)
c. \(\int x\left(x^2+1\right)^{10}dx=\dfrac{1}{2}\int\left(x^2+1\right)^{10}d\left(x^2+1\right)=\dfrac{\left(x^2+1\right)^{11}}{22}+C\)
d. \(\int x\left(2-x\right)^5dx\)
Đặt \(\left\{{}\begin{matrix}u=x\\dv=\left(2-x\right)^5dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=dx\\v=-\dfrac{1}{6}\left(2-x\right)^6\end{matrix}\right.\)
\(\Rightarrow I=-\dfrac{x\left(2-x\right)^6}{6}+\dfrac{1}{6}\int\limits\left(2-x\right)^6dx=-\dfrac{x\left(2-x\right)^6}{6}+\dfrac{1}{42}\left(x-2\right)^7+C\)
e.
\(I=\int\dfrac{2x+1}{\left(x-3\right)^3}dx=\int\dfrac{2\left(x-3\right)+6}{\left(x-3\right)^3}dx=2\int\dfrac{dx}{\left(x-3\right)^2}+6\int\dfrac{dx}{\left(x-3\right)^3}\)
\(=-\dfrac{2}{x-3}-\dfrac{3}{\left(x-3\right)^2}+C\)
f.
\(I=\int\dfrac{\left(2x-1\right)^4}{\left(x+3\right)^6}dx=\int\dfrac{\left[2\left(x+3\right)-7\right]^4}{\left(x+3\right)^6}dx=\int\left(2-\dfrac{7}{x+3}\right)^4.\dfrac{1}{\left(x+3\right)^2}dx\)
Đặt \(2-\dfrac{7}{x+3}=u\Rightarrow du=\dfrac{7}{\left(x+3\right)^2}dx\Rightarrow\dfrac{1}{\left(x+3\right)^2}dx=\dfrac{1}{7}du\)
\(\Rightarrow I=\dfrac{1}{7}\int u^4du==\dfrac{1}{35}.u^5+C=\dfrac{1}{35}\left(2-\dfrac{7}{x+3}\right)^5+C\)
g.
\(I=\int x\sqrt{1-x}dx\)
Đặt \(\sqrt{1-x}=u\Rightarrow x=1-u^2\Rightarrow dx=-2u.du\)
\(I=\int\left(1-u^2\right).u.\left(-2u.du\right)=\int\left(2u^4-2u^2\right)du=\dfrac{2}{5}u^5-\dfrac{2}{3}u^3+C\)
\(=\dfrac{2}{5}\sqrt{\left(1-x\right)^5}-\dfrac{2}{3}\sqrt{\left(1-x\right)^3}+C\)
h.
\(I=\int\dfrac{\left(1-2e^x\right)e^x}{e^x+1}dx\)
Đặt \(e^x+1=u\Rightarrow e^xdx=du\)
\(I=\int\dfrac{\left(1-2\left(u-1\right)\right)}{u}du=\int\dfrac{3-2u}{u}du=\int\left(\dfrac{3}{u}-2\right)du=3lnu-2u+C\)
\(=3ln\left(e^x+1\right)-2\left(e^x+1\right)+C=3ln\left(e^x+1\right)-2e^x+C\)
i.
Đặt \(lnx=u\Rightarrow\dfrac{dx}{x}=du\)
\(I=\int\left(u^2-3u\right)du=\dfrac{u^3}{3}-\dfrac{3u^2}{2}+C=\dfrac{ln^3x}{3}=\dfrac{3ln^2x}{2}+C\)
Cho F(x) = 1/x là một nguyên hàm của x^2f(x). Tìm nguyên hàm của f'(x)x^3lnx
Giải giúp em với mấy anh ;-;
Ta có: \(F\left(x\right)=\dfrac{1}{x}=\int x^2f\left(x\right)dx\)
Hay \(F'\left(x\right)=-\dfrac{1}{x^2}=x^2f\left(x\right)\Rightarrow f\left(x\right)=\dfrac{-1}{x^4}\)
Có: \(f'\left(x\right)=\dfrac{4}{x^3}\) \(\Rightarrow I=\int f'\left(x\right)x^3lnxdx=\int\dfrac{4}{x^3}x^3lnxdx=4\int lnxdx\)
Đặt: \(\left\{{}\begin{matrix}u=lnx\\dv=dx\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{1}{x}dx\\v=x\end{matrix}\right.\)
\(\Rightarrow I=\text{4}(xlnx-\int x\dfrac{1}{x}dx)=4x\left(lnx-1\right)\)
Tính nguyên hàm
\(\int e^x\left(2+\dfrac{e^{-x}}{x}\right)dx=\int\left(2e^x+\dfrac{1}{x}\right)dx=2e^x+ln\left|x\right|+C\)
Tính nguyên hàm
\(I=\int\left(\dfrac{1}{2}+\dfrac{1}{2}cos4x\right)=\dfrac{1}{2}x+\dfrac{1}{8}sin4x+C\)
\(\int\dfrac{dx}{x^2+4}\)
\(\int\dfrac{dx}{x^2+4}=\dfrac{1}{2}arctan\left(\dfrac{x}{2}\right)+C\)
Tính nguyên hàm
\(\int\dfrac{1}{\left(4-2x\right)^5}dx\)
\(=-\dfrac{1}{2}\int\dfrac{d\left(4-2x\right)}{\left(4-2x\right)^5}=\dfrac{1}{8.\left(4-2x\right)^4}+C\)
Cho em xin lời giải bài 3 và bài 4 với ạ
Chỉ thấy bài 5 với 6:
5.
\(f'\left(x\right)+2f\left(x\right)=0\Leftrightarrow f'\left(x\right)=-2f\left(x\right)\Leftrightarrow\dfrac{f'\left(x\right)}{f\left(x\right)}=-2\)
Lấy nguyên hàm 2 vế:
\(\int\dfrac{f'\left(x\right)}{f\left(x\right)}dx=\int-2dx\Rightarrow ln\left(f\left(x\right)\right)=-2x+C\)
Thay \(x=1\Rightarrow0=-2+C\Rightarrow C=2\)
\(\Rightarrow ln\left(f\left(x\right)\right)=-2x+2\Rightarrow f\left(x\right)=e^{-2x+2}\)
\(\Rightarrow f\left(-1\right)=e^4\)
6.
\(f\left(x\right)+x.f'\left(x\right)=2x+1\)
\(\Leftrightarrow x'.f\left(x\right)+x.f'\left(x\right)=2x+1\)
\(\Leftrightarrow\left[x.f\left(x\right)\right]'=2x+1\)
Lấy nguyên hàm 2 vế:
\(\int\left[x.f\left(x\right)\right]'dx=\int\left(2x+1\right)dx\)
\(\Rightarrow x.f\left(x\right)=x^2+x+C\)
Thay \(x=1\Rightarrow1.f\left(1\right)=1+1+C\Rightarrow C=1\)
\(\Rightarrow f\left(x\right)=\dfrac{x^2+x+1}{x}\)
\(\Rightarrow f\left(2\right)=\dfrac{7}{2}\)
Ủa sao đề khác rồi:
3.
\(f'\left(x\right)=-e^x.f^2\left(x\right)\Leftrightarrow\dfrac{f'\left(x\right)}{f^2\left(x\right)}=-e^x\)
Lấy nguyên hàm 2 vế:
\(\int\dfrac{f'\left(x\right)}{f^2\left(x\right)}dx=\int-e^xdx\)
\(\Rightarrow\dfrac{1}{f\left(x\right)}=e^x+C\)
Thay \(x=0\Rightarrow2=1+C\Rightarrow C=1\)
\(\Rightarrow\dfrac{1}{f\left(x\right)}=e^x+1\Rightarrow f\left(x\right)=\dfrac{1}{e^x+1}\)
\(\Rightarrow f\left(ln2\right)=\dfrac{1}{3}\)
\(dx\)
Tìm: I = \(\int x^2\left(2-3x^2\right)^8dx\)
Nguyên hàm này có thể coi là không tính được (cách tính duy nhất là khai triển nhị thức Newton của \(\left(2-3x^2\right)^8\) ra thành dạng đa thức sau đó tính nguyên hàm, nhưng chắc ko ai cho đề như vậy cả)
I=\(\int x^2\left(2-3x^2\right)^8dx\)
đặt y = \(2-3x^2\)\(\Rightarrow\left\{{}\begin{matrix}dy=-6xdx\\x^2=\dfrac{2-y}{3}\end{matrix}\right.\)\(\Leftrightarrow x^2\left(2-3x^2\right)^8=\left(\dfrac{2-y}{3}\right).y^8=\dfrac{1}{3}\left(2y^8-y^9\right)\)
vậy thì :
\(I=\int x^2\left(2-3x^2\right)^8dx=\dfrac{1}{3}\left(2\int y^8dy-\int y^9dy\right)\)
\(=\dfrac{2}{27}y^9-\dfrac{1}{30}y^{10}+C\)
\(=\dfrac{2}{27}\left(2-3x^2\right)^9-\dfrac{1}{30}\left(2-3x^2\right)^{10}+C\)
Tính :
\(I_1=\int\dfrac{x^2-2x+2}{\sqrt{x^2-2x}}dx\)
\(I_2=\dfrac{dx}{\left(x-1\right)\sqrt{x^2-2x+2}}\)
\(I_3=\dfrac{dx}{\left(2x+1\right)\sqrt{x^2-2x+2}}\)