\(I=\int\dfrac{x.e^x}{\left(x+1\right)^2}dx\)
Những câu hỏi liên quan
1, I intlimits^1_0dfrac{2x+1}{x^2+x+1}dx
2,intlimits^{dfrac{1}{2}}_0dfrac{5xdx}{left(1-x^2right)^3}
3, intlimits^1_0dfrac{2x}{left(x+1right)^3}dx
4, intlimits^1_0dfrac{4x-2}{left(x^2+1right)left(x+2right)}dx
5, intlimits^1_0dfrac{x^2dx}{x^6-9}
6, intlimits^2_1dfrac{2x-1}{x^2left(x+1right)}dx
Đọc tiếp
1, I = \(\int\limits^1_0\dfrac{2x+1}{x^2+x+1}dx\)
2,\(\int\limits^{\dfrac{1}{2}}_0\dfrac{5xdx}{\left(1-x^2\right)^3}\)
3, \(\int\limits^1_0\dfrac{2x}{\left(x+1\right)^3}dx\)
4, \(\int\limits^1_0\dfrac{4x-2}{\left(x^2+1\right)\left(x+2\right)}dx\)
5, \(\int\limits^1_0\dfrac{x^2dx}{x^6-9}\)
6, \(\int\limits^2_1\dfrac{2x-1}{x^2\left(x+1\right)}dx\)
1/ \(I=\int\limits^1_0\dfrac{2x+1}{x^2+x+1}dx=\int\limits^1_0\dfrac{d\left(x^2+x+1\right)}{x^2+x+1}=ln\left|x^2+x+1\right||^1_0=ln3\)
2/ \(\int\limits^{\dfrac{1}{2}}_0\dfrac{5x}{\left(1-x^2\right)^3}dx=-\dfrac{5}{2}\int\limits^{\dfrac{1}{2}}_0\dfrac{d\left(1-x^2\right)}{\left(1-x^2\right)^3}=\dfrac{5}{4}\dfrac{1}{\left(1-x^2\right)^2}|^{\dfrac{1}{2}}_0=\dfrac{35}{36}\)
3/ \(\int\limits^1_0\dfrac{2x}{\left(x+1\right)^3}dx\Rightarrow\) đặt \(x+1=t\Rightarrow x=t-1\Rightarrow dx=dt;\left\{{}\begin{matrix}x=0\Rightarrow t=1\\x=1\Rightarrow t=2\end{matrix}\right.\)
\(I=\int\limits^2_1\dfrac{2\left(t-1\right)dt}{t^3}=\int\limits^2_1\left(\dfrac{2}{t^2}-\dfrac{2}{t^3}\right)dt=\left(\dfrac{-2}{t}+\dfrac{1}{t^2}\right)|^2_1=\dfrac{1}{4}\)
4/ \(\int\limits^1_0\dfrac{4x-2}{\left(x^2+1\right)\left(x+2\right)}dx\)
Kĩ thuật chung là tách và sử dụng hệ số bất định như sau:
\(\dfrac{4x-2}{\left(x^2+1\right)\left(x+2\right)}=\dfrac{ax+b}{x^2+1}+\dfrac{c}{x+2}=\dfrac{\left(a+c\right)x^2+\left(2a+b\right)x+2b+c}{\left(x^2+1\right)\left(x+2\right)}\)
\(\Rightarrow\left\{{}\begin{matrix}a+c=0\\2a+b=4\\2b+c=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b=0\\a=-c=2\end{matrix}\right.\)
\(\Rightarrow I=\int\limits^1_0\left(\dfrac{2x}{x^2+1}-\dfrac{2}{x+2}\right)dx=\int\limits^1_0\dfrac{d\left(x^2+1\right)}{x^2+1}-2\int\limits^1_0\dfrac{d\left(x+2\right)}{x+2}=ln\dfrac{8}{9}\)
5/ \(\int\limits^1_0\dfrac{x^2dx}{x^6-9}\Rightarrow\) đặt \(x^3=t\Rightarrow3x^2dx=dt\Rightarrow x^2dx=\dfrac{1}{3}dt;\left\{{}\begin{matrix}x=0\Rightarrow t=0\\x=1\Rightarrow t=1\end{matrix}\right.\)
\(I=\dfrac{1}{3}\int\limits^1_0\dfrac{dt}{t^2-9}=\dfrac{1}{18}\int\limits^1_0\left(\dfrac{1}{t-3}-\dfrac{1}{t+3}\right)dt=\dfrac{1}{18}ln\left|\dfrac{t-3}{t+3}\right||^1_0=-\dfrac{1}{18}ln2\)
6/ Tương tự câu 4, sử dụng hệ số bất định ta tách được:
\(\int\limits^2_1\dfrac{2x-1}{x^2\left(x+1\right)}dx=\int\limits^2_1\left(\dfrac{3x-1}{x^2}-\dfrac{3}{x+1}\right)dx=\int\limits^2_1\left(\dfrac{3}{x}-\dfrac{1}{x^2}-\dfrac{3}{x+1}\right)dx\)
\(=\left(3ln\left|\dfrac{x}{x+1}\right|+\dfrac{1}{x}\right)|^2_1=3ln\dfrac{4}{3}-\dfrac{1}{2}\)
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Tính :
a) intleft(2-xright)sin xdx
b) intdfrac{left(x+1right)^2}{sqrt{x}}dx
c) intdfrac{3^{3x}+1}{e^x+1}dx
d) intdfrac{1}{left(sin x+cos xright)^2}dx
e) intdfrac{1}{sqrt{1+x}+sqrt{x}}dx
g) intdfrac{1}{left(1+xright)left(2-xright)}dx
Đọc tiếp
Tính :
a) \(\int\left(2-x\right)\sin xdx\)
b) \(\int\dfrac{\left(x+1\right)^2}{\sqrt{x}}dx\)
c) \(\int\dfrac{3^{3x}+1}{e^x+1}dx\)
d) \(\int\dfrac{1}{\left(\sin x+\cos x\right)^2}dx\)
e) \(\int\dfrac{1}{\sqrt{1+x}+\sqrt{x}}dx\)
g) \(\int\dfrac{1}{\left(1+x\right)\left(2-x\right)}dx\)
Áp dụng phương pháp tính tích phân, hãy tính các tích phân sau :
a) intlimits^{dfrac{pi}{2}}_0xcos2xdx
b) intlimits^{ln2}_0xe^{-2x}dx
c) intlimits^1_0lnleft(2x+1right)dx
d) intlimits^3_2left|lnleft(x-1right)-lnleft(x+1right)right|dx
e) intlimits^2_{dfrac{1}{2}}left(1+x-dfrac{1}{x}right)e^{x+dfrac{1}{x}}dx
g) intlimits^{dfrac{pi}{2}}_0xcos xsin^2xdx
h) intlimits^1_0dfrac{xe^x}{left(1+xright)^2}dx
i) intlimits^e_1dfrac{1+xln x}{x}e^xdx
Đọc tiếp
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) \(\int\limits^e_1\dfrac{1+x\ln x}{x}e^xdx\)
\(\int\dfrac{1}{cosx.cos\left(x+\dfrac{\pi}{4}\right)}dx\)
\(\int\dfrac{1}{x^3\left(1+x^2\right)}dx=\dfrac{a}{x^2}+blnx+cln\left(1+x^2\right).S=a+b+c=?\)
\(\int\dfrac{5-3x}{\left(x^2-5x+6\right)\left(x^2-2x+1\right)}dx=\dfrac{a}{x-1}-ln\left(\dfrac{x-b}{x-c}\right)+C.P=2a+b\)
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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Tính các nguyên hàm sau :
a) int xleft(3-xright)^5dx
b) intleft(2^x-3^xright)^2dx
c) int xsqrt{2-5x}dx
d) intdfrac{lnleft(cos xright)}{cos^2x}dx
e) intdfrac{x}{sin^2x}dx
intdfrac{x+1}{left(x-2right)left(x+3right)}dx
h) intdfrac{1}{1-sqrt{x}}dx
i) intsin3xcos2xdx
k) intdfrac{sin^3x}{cos^2x}dx
l) intdfrac{sin xcos x}{sqrt{a^2sin^2x+b^2cos^2x}}dx (a^2ne b^2)
Đọc tiếp
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: \(I=\int\dfrac{dx}{\sqrt{\left(x+1\right)\left(x+2\right)}}\)
\(I=\int\dfrac{dx}{\sqrt{\left(x+\dfrac{3}{2}\right)^2+\dfrac{1}{4}}}\)
Đặt \(x+\dfrac{3}{2}=\dfrac{1}{2}tanu\Rightarrow dx=\dfrac{1}{2cos^2u}du\)
\(I=\int\dfrac{1}{\dfrac{1}{2}\sqrt{tan^2u+1}}.\dfrac{1}{2.cos^2u}du=\int\dfrac{1}{cosu}du=\int\dfrac{1}{1-sin^2u}d\left(sinu\right)\)
\(=\dfrac{1}{2}ln\left|\dfrac{1+sinu}{1-sinu}\right|+C=ln\left(\dfrac{1+sinu}{cosu}\right)+C=ln\left(\dfrac{1}{cosu}+tanu\right)+C\)
Chú ý: \(\dfrac{1}{cosu}=\sqrt{\dfrac{1}{cos^2u}}=\sqrt{1+tan^2u}=\sqrt{1+\left(2x+3\right)^2}=2\sqrt{x^2+3x+2}\)
Do đó: \(I=ln\left(2x+3+2\sqrt{x^2+3x+2}\right)+C\)
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Ủa giờ mới để ý tách biểu thức sai, \(x^2+3x+2=\left(x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}\) mới đúng
Vậy làm cách khác:
Đặt \(\sqrt{\left(x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}}=-\left(x+\dfrac{3}{2}\right)+t\)
\(\Rightarrow\left(x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}=\left(x+\dfrac{3}{2}\right)^2-2t\left(x+\dfrac{3}{2}\right)+t^2\)
\(\Rightarrow x+\dfrac{3}{2}=\dfrac{\dfrac{1}{4}+t^2}{2t}=\dfrac{1}{8t}+\dfrac{t}{2}\)
\(\Rightarrow dx=\left(-\dfrac{1}{8t^2}+\dfrac{1}{2}\right)dt=\left(\dfrac{4t^2-1}{8t^2}\right)dt\)
Lại có: \(\sqrt{x^2+3x+2}=-\left(x+\dfrac{3}{2}\right)+t=-\dfrac{1}{8t}-\dfrac{t}{2}+t=\dfrac{t}{2}-\dfrac{1}{8t}=\dfrac{4t^2-1}{8t}\)
\(\Rightarrow\dfrac{1}{\sqrt{x^2+3x+2}}=\dfrac{8t}{4t^2-1}\)
Do đó:
\(I=\int\dfrac{8t}{4t^2-1}.\dfrac{4t^2-1}{8t^2}=\int\dfrac{1}{t}dt=ln\left|t\right|+C\)
\(=ln\left|\sqrt{x^2+3x+2}+\left(x+\dfrac{3}{2}\right)\right|+C=ln\left|2\sqrt{x^2+3x+2}+2x+3\right|+C\)
Lần này chắc ko nhầm nữa :D
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19 tháng 1 lúc 20:06
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 \(\left\{{}\begin{matrix}u=lnx\\dv=xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=\dfrac{1}{2}x^2\end{matrix}\right.\)
\(\Rightarrow I=\dfrac{1}{2}x^2.lnx-\dfrac{1}{2}\int xdx=\dfrac{1}{2}x^2.lnx-\dfrac{1}{4}x^2+C\)
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Tính các tích phân sau :
a) intlimits^1_0left(y^3+3y^2-2right)dy
b) intlimits^4_1left(t+dfrac{1}{sqrt{t}}-dfrac{1}{t^2}right)dt
c) intlimits^{dfrac{pi}{2}}_0left(2cos x-sin2xright)dx
d) intlimits^1_0left(3^s-2^sright)^2ds
e) intlimits^{dfrac{pi}{3}}_0cos3xdx+intlimits^{dfrac{3pi}{2}}_0cos3xdx+intlimits^{dfrac{5pi}{2}}_{dfrac{3pi}{2}}cos3xdx
g) intlimits^3_0left|x^2-x-2right|dx
h) intlimits^{dfrac{5pi}{4}}_{pi}dfrac{sin x-cos x}{sqrt{1+sin2x}}dx
i) intlimits^4_0dfrac{4x-1}{sqrt{2x+1}+2}dx
Đọc tiếp
Tính các tích phân sau :
a) \(\int\limits^1_0\left(y^3+3y^2-2\right)dy\)
b) \(\int\limits^4_1\left(t+\dfrac{1}{\sqrt{t}}-\dfrac{1}{t^2}\right)dt\)
c) \(\int\limits^{\dfrac{\pi}{2}}_0\left(2\cos x-\sin2x\right)dx\)
d) \(\int\limits^1_0\left(3^s-2^s\right)^2ds\)
e) \(\int\limits^{\dfrac{\pi}{3}}_0\cos3xdx+\int\limits^{\dfrac{3\pi}{2}}_0\cos3xdx+\int\limits^{\dfrac{5\pi}{2}}_{\dfrac{3\pi}{2}}\cos3xdx\)
g) \(\int\limits^3_0\left|x^2-x-2\right|dx\)
h) \(\int\limits^{\dfrac{5\pi}{4}}_{\pi}\dfrac{\sin x-\cos x}{\sqrt{1+\sin2x}}dx\)
i) \(\int\limits^4_0\dfrac{4x-1}{\sqrt{2x+1}+2}dx\)
Câu nào mình biết thì mình làm nha.
1) Đổi thành \(\dfrac{y^4}{4}+y^3-2y\) rồi thế số.KQ là \(\dfrac{-3}{4}\)
2) Biến đổi thành \(\dfrac{t^2}{2}+2\sqrt{t}+\dfrac{1}{t}\) và thế số.KQ là \(\dfrac{35}{4}\)
3) Biến đổi thành 2sinx + cos(2x)/2 và thế số.KQ là 1
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Tính các tích phân sau :
a) intlimits^{dfrac{1}{2}}_{-dfrac{1}{2}}sqrt[3]{left(1-xright)^2dx}
b) intlimits^{dfrac{pi}{2}}_0sinleft(dfrac{pi}{4}-xright)dx
c) intlimits^2_{dfrac{1}{2}}dfrac{1}{xleft(x+1right)}dx
d) intlimits^2_0xleft(x+1right)^2dx
e) intlimits^2_{dfrac{1}{2}}dfrac{1-3x}{left(x+1right)^2}dx
g) intlimits^{dfrac{pi}{2}}_{-dfrac{pi}{2}}sin3xcos5xdx
Đọc tiếp
Tính các tích phân sau :
a) \(\int\limits^{\dfrac{1}{2}}_{-\dfrac{1}{2}}\sqrt[3]{\left(1-x\right)^2dx}\)
b) \(\int\limits^{\dfrac{\pi}{2}}_0\sin\left(\dfrac{\pi}{4}-x\right)dx\)
c) \(\int\limits^2_{\dfrac{1}{2}}\dfrac{1}{x\left(x+1\right)}dx\)
d) \(\int\limits^2_0x\left(x+1\right)^2dx\)
e) \(\int\limits^2_{\dfrac{1}{2}}\dfrac{1-3x}{\left(x+1\right)^2}dx\)
g) \(\int\limits^{\dfrac{\pi}{2}}_{-\dfrac{\pi}{2}}\sin3x\cos5xdx\)
a) =
=
b) =
=
=
c)=
d)=
=
e)=
=
g)Ta có f(x) = sin3xcos5x là hàm số lẻ.
Vì f(-x) = sin(-3x)cos(-5x) = -sin3xcos5x = f(-x) nên:
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