Tìm:
a) \(\int x^{\dfrac{3}{5}}dx;\) b) \(\int\dfrac{1}{\sqrt[4]{x^3}}dx.\)
Những câu hỏi liên quan
Tính nguyên hàm của:1, intdfrac{x^3}{x-2}dx2, intdfrac{dx}{xsqrt{x^2+1}}3, int(dfrac{5}{x}+sqrt{x^3})dx4, intdfrac{xsqrt{x}+sqrt{x}}{x^2}dx5, intdfrac{dx}{sqrt{1-x^2}}
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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. \(\int\dfrac{dx}{\sqrt{1-x^2}}=arcsin\left(x\right)+C\)
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1, \(\int\dfrac{x^3dx}{\left(x^8-4\right)^2}\)
2, \(\int\dfrac{2x+1}{x^4+2x^2+3x^2+2x-3}dx\)
3, \(\int\dfrac{sin\sqrt[3]{x}}{\sqrt[3]{x^2}}dx\)
4, \(\int\dfrac{dx}{sin^2x+2cos^2x}\)
5, \(\int\dfrac{sinx+cosx}{3+sin2x}dx\)
\(I=\int\dfrac{x^3dx}{\left(x^8-4\right)^2}\)
Đặt \(x^4=t\Rightarrow x^3dx=\dfrac{1}{4}dt\Rightarrow I=\dfrac{1}{4}\int\dfrac{dt}{\left(t^2-2\right)^2}=\dfrac{1}{4}\int\dfrac{dt}{\left(t-\sqrt{2}\right)^2\left(t+\sqrt{2}\right)^2}\)
\(=\dfrac{1}{32}\int\left(\dfrac{1}{t-\sqrt{2}}-\dfrac{1}{t+\sqrt{2}}\right)^2dt=\dfrac{1}{32}\int\left(\dfrac{1}{\left(t-\sqrt{2}\right)^2}+\dfrac{1}{\left(t+\sqrt{2}\right)^2}-\dfrac{2}{\left(t+\sqrt{2}\right)\left(t-\sqrt{2}\right)}\right)dt\)
\(=\dfrac{1}{32}\int\left(\dfrac{1}{\left(t-\sqrt{2}\right)^2}+\dfrac{1}{\left(t+\sqrt{2}\right)^2}-\dfrac{1}{\sqrt{2}}\left(\dfrac{1}{t-\sqrt{2}}-\dfrac{1}{t+\sqrt{2}}\right)\right)dt\)
\(=\dfrac{1}{32}\left(\dfrac{-1}{t-\sqrt{2}}-\dfrac{1}{t+\sqrt{2}}-\dfrac{1}{\sqrt{2}}ln\left|\dfrac{t-\sqrt{2}}{t+\sqrt{2}}\right|\right)+C\)
\(=\dfrac{1}{32}\left(\dfrac{-1}{x^4-\sqrt{2}}-\dfrac{1}{x^4+\sqrt{2}}-\dfrac{1}{\sqrt{2}}ln\left|\dfrac{x^4-\sqrt{2}}{x^4+\sqrt{2}}\right|\right)+C\)
2/ \(I=\int\dfrac{\left(2x+1\right)dx}{\left(x^2+x-1\right)\left(x^2+x+3\right)}=\dfrac{1}{4}\int\left(\dfrac{1}{x^2+x-1}-\dfrac{1}{x^2+x+3}\right)\left(2x+1\right)dx\)
\(=\dfrac{1}{4}\int\left(\dfrac{2x+1}{x^2+x-1}-\dfrac{2x+1}{x^2+x+3}\right)dx\)
\(=\dfrac{1}{4}\left(\int\dfrac{d\left(x^2+x-1\right)}{x^2+x-1}-\int\dfrac{d\left(x^2+x+3\right)}{x^2+x+3}\right)\)
\(=\dfrac{1}{4}ln\left|\dfrac{x^2+x-1}{x^2+x+3}\right|+C\)
3/ Đặt \(\sqrt[3]{x}=t\Rightarrow x=t^3\Rightarrow dx=3t^2dt\)
\(\Rightarrow I=\int\dfrac{3t^2.sint.dt}{t^2}=3\int sint.dt=-3cost+C=-3cos\left(\sqrt[3]{x}\right)+C\)
4/ \(I=\int\dfrac{dx}{1+cos^2x}=\int\dfrac{\dfrac{1}{cos^2x}dx}{\dfrac{1}{cos^2x}+1}\)
Đặt \(t=tanx\Rightarrow\left\{{}\begin{matrix}dt=\dfrac{1}{cos^2x}dx\\\dfrac{1}{cos^2x}=1+tan^2x=1+t^2\end{matrix}\right.\)
\(\Rightarrow I=\int\dfrac{dt}{1+t^2+1}=\int\dfrac{dt}{t^2+2}=\dfrac{1}{2}\int\dfrac{dt}{\left(\dfrac{t}{\sqrt{2}}\right)^2+1}\)
\(=\dfrac{1}{2}.\sqrt{2}.arctan\left(\dfrac{t}{\sqrt{2}}\right)+C=\dfrac{1}{\sqrt{2}}arctan\left(\dfrac{tanx}{\sqrt{2}}\right)+C\)
5/ \(I=\int\dfrac{sinx+cosx}{4+2sinx.cosx-sin^2x-cos^2x}dx=\int\dfrac{sinx+cosx}{4-\left(sinx-cosx\right)^2}dx\)
Đặt \(sinx-cosx=t\Rightarrow\left(cosx+sinx\right)dx=dt\)
\(\Rightarrow I=\int\dfrac{dt}{4-t^2}=-\int\dfrac{dt}{\left(t-2\right)\left(t+2\right)}=\dfrac{1}{4}\int\left(\dfrac{1}{t+2}-\dfrac{1}{t-2}\right)dt\)
\(=\dfrac{1}{4}ln\left|\dfrac{t+2}{t-2}\right|+C=\dfrac{1}{4}ln\left|\dfrac{sinx-cosx+2}{sinx-cosx-2}\right|+C\)
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Ơ bài 1 nhầm số 4 thành số 2 rồi, bạn sửa lại 1 chút nhé :D
Còn 1 cách làm khác nữa là lượng giác hóa
Đặt \(x^4=2sint\Rightarrow x^3dx=\dfrac{1}{2}cost.dt\)
\(\Rightarrow I=\dfrac{1}{2}\int\dfrac{cost.dt}{\left(4sin^2t-4\right)^2}=\dfrac{1}{32}\int\dfrac{cost.dt}{cos^4t}=\dfrac{1}{32}\int\dfrac{dt}{cos^3t}\)
Đặt \(\left\{{}\begin{matrix}u=\dfrac{1}{cost}\\dv=\dfrac{dt}{cos^2t}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{sint.dt}{cos^2t}\\v=tant\end{matrix}\right.\)
\(\Rightarrow32I=\dfrac{tant}{cost}-\int\dfrac{tant.sint.dt}{cos^2t}=\dfrac{sint}{cos^2t}-\int\dfrac{sin^2t.dt}{cos^3t}\)
\(=\dfrac{sint}{1-sin^2t}-\int\dfrac{1-cos^2t}{cos^3t}dt=\dfrac{sint}{1-sin^2t}-\int\dfrac{dt}{cos^3t}+\int\dfrac{1}{cosx}dx\)
Chú ý rằng \(\int\dfrac{dt}{cos^3t}=32I\)
\(\Rightarrow32I=\dfrac{sint}{1-sin^2t}-32I+\int\dfrac{cost.dt}{cos^2t}\)
\(\Rightarrow64I=\dfrac{sint}{1-sin^2t}-\int\dfrac{d\left(sint\right)}{sin^2t-1}=\dfrac{sint}{1-sin^2t}-\dfrac{1}{2}ln\left|\dfrac{sint-1}{sint+1}\right|+C\)
\(\Rightarrow I=\dfrac{1}{64}\left(\dfrac{2x^4}{4-x^8}-\dfrac{1}{2}ln\left|\dfrac{x^4-2}{x^4+2}\right|\right)+C\)
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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
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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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1.\(\int\dfrac{\sin2x}{\cos^4x-4}dx\)
2.\(\int\sqrt{1-x^2}dx\)
3.\(\int\dfrac{xdx}{\sqrt{1+x^4}}dx\)
giúp mình với mn
mọi người tính giúp em mấy cái nguyên hàm với :
1, intdfrac{x^2-1}{x^4+1}dx
2, intdfrac{dx}{3x^2-3}
3, intdfrac{dx}{1+sinx}
4, intdfrac{sinxdx}{sqrt{cos2x}}
5, intdfrac{xdx}{sqrt[3]{1-3x}}
6, int2^{2x}.3^{3x}.4^xdx
7,intdfrac{dx}{xlnx.lnleft(lnxright)}
8, intdfrac{dx}{sinx.cos^3x}
9, intdfrac{cos2xdx}{sin^2x.cos^2x}
10, intdfrac{dx}{sqrt{2x}-sqrt{2x+1}}
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mọi người tính giúp em mấy cái nguyên hàm với :
1, \(\int\dfrac{x^2-1}{x^4+1}dx\)
2, \(\int\dfrac{dx}{3x^2-3}\)
3, \(\int\dfrac{dx}{1+sinx}\)
4, \(\int\dfrac{sinxdx}{\sqrt{cos2x}}\)
5, \(\int\dfrac{xdx}{\sqrt[3]{1-3x}}\)
6, \(\int2^{2x}.3^{3x}.4^xdx\)
7,\(\int\dfrac{dx}{xlnx.ln\left(lnx\right)}\)
8, \(\int\dfrac{dx}{sinx.cos^3x}\)
9, \(\int\dfrac{cos2xdx}{sin^2x.cos^2x}\)
10, \(\int\dfrac{dx}{\sqrt{2x}-\sqrt{2x+1}}\)
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
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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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1) \(\int\limits^2_1\dfrac{\sqrt{x^2-1}}{x}dx\)
2) \(\int x.\sqrt{1-x^4}dx\)
3)\(\int\dfrac{1}{x^2.\sqrt{25-x^2}}dx\)
4) \(\int\dfrac{\sqrt{x^2-9}}{x^3}dx\)
Tính các tích phân sau :
a) intlimits^4_{-2}left(dfrac{x-2}{x+3}right)^2dx (đặt tx+3 )
b) intlimits^6_{-4}left|x+3right|-left|x-4right|dx
c) intlimits^2_{-3}dfrac{dx}{sqrt{x+7}+3} (đặt tsqrt{x+7} hoặc tsqrt{x+7}+3 )
d) intlimits^{dfrac{pi}{2}}_0dfrac{cos x}{1+4sin x}dx
e) intlimits^2_1dfrac{x^9}{x^{10}+4x^5+4}dx (đặt tx^5 )
g) intlimits^3_0left(x+2right)e^{2x}dx
h) intlimits^5_2dfrac{sqrt{4+x}}{x}dx (đặt tsqrt{4+x} )
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Tính các tích phân sau :
a) \(\int\limits^4_{-2}\left(\dfrac{x-2}{x+3}\right)^2dx\) (đặt \(t=x+3\) )
b) \(\int\limits^6_{-4}\left|x+3\right|-\left|x-4\right|dx\)
c) \(\int\limits^2_{-3}\dfrac{dx}{\sqrt{x+7}+3}\) (đặt \(t=\sqrt{x+7}\) hoặc \(t=\sqrt{x+7}+3\) )
d) \(\int\limits^{\dfrac{\pi}{2}}_0\dfrac{\cos x}{1+4\sin x}dx\)
e) \(\int\limits^2_1\dfrac{x^9}{x^{10}+4x^5+4}dx\) (đặt \(t=x^5\) )
g) \(\int\limits^3_0\left(x+2\right)e^{2x}dx\)
h) \(\int\limits^5_2\dfrac{\sqrt{4+x}}{x}dx\) (đặt \(t=\sqrt{4+x}\) )
1) Tinh nguyen ham
a) A = \(\int\dfrac{x}{\sqrt{x+2}}.dx\) b) B = \(\int\dfrac{sinx+cosx}{\sqrt[3]{1-sin2x}}.dx\)
\(A=\int \frac{x}{\sqrt{x+2}}dx \\ = \int \frac{x+2-2}{\sqrt{x+2}}dx \\ = \int \sqrt{x+2}-2\frac{1}{\sqrt{x+2}}dx \\ = \frac{2}{3}(x+2)^{\frac{3}{2}}-4\sqrt{x+2}+C\)
\(B=\int \frac{sinx+cosx}{\sqrt[3]{1-sin2x}}dx \\ x=\frac{\pi}{4}-u, dx=-du \\ =- \int \frac{sin(\frac{\pi}{4}-u)+cos(\frac{\pi}{4}-u)}{\sqrt[3]{1-sin(\frac{\pi}{2}-2u)}}du \\ = - \int \frac{\frac{1}{\sqrt2}cosu+\frac{1}{\sqrt2}sinu+\frac{1}{\sqrt2}cosu-\frac{1}{\sqrt2}sinu}{\sqrt[3]{1-cos2u}}du \\ = -\int \frac{\frac{2}{\sqrt2}cosu}{\sqrt[3]{1-cos2u}}du \\ = -\sqrt2 \int \frac{cosu}{\sqrt[3]{1-cos^2u+sin^2u}}du \\ = -\sqrt2 \int \frac{cosu}{\sqrt[3]{2sin^2u}}du \\ v=sinu, dv=cosudu \\ = -\sqrt2 \int \frac{1}{\sqrt[3]{2v^2}}dv \\ = -\frac{\sqrt2}{\sqrt[3]2} \int v^{-\frac{2}{3}}dv \\ = -\frac{\sqrt2}{\sqrt[3]2} 3v^\frac{1}{3}+C \\ = -\frac{\sqrt2}{\sqrt[3]2} 3\sqrt[3]{sin(\frac{\pi}{4}-x)}+C \)
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\(\int\dfrac{1}{cosx.cos\left(x+\dfrac{\pi}{4}\right)}dx\)
\(\int\dfrac{1}{x^3\left(1+x^2\right)}dx=\dfrac{a}{x^2}+blnx+cln\left(1+x^2\right).S=a+b+c=?\)
\(\int\dfrac{5-3x}{\left(x^2-5x+6\right)\left(x^2-2x+1\right)}dx=\dfrac{a}{x-1}-ln\left(\dfrac{x-b}{x-c}\right)+C.P=2a+b\)
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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