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Câu 1: A=\(\int\dfrac{2SINX+COSX}{3SINX+2COSX}DX\)
Câu 2: \(I=\int\dfrac{X^3}{X^4+3X^2+2}\)
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NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 2 2019
\(I=\int\dfrac{x^3dx}{\left(x^8-4\right)^2}\)
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 \(sinx-cosx=t\Rightarrow\left(cosx+sinx\right)dx=dt\)
\(\Rightarrow I=\int\dfrac{dt}{4-t^2}=-\int\dfrac{dt}{\left(t-2\right)\left(t+2\right)}=\dfrac{1}{4}\int\left(\dfrac{1}{t+2}-\dfrac{1}{t-2}\right)dt\)
\(=\dfrac{1}{4}ln\left|\dfrac{t+2}{t-2}\right|+C=\dfrac{1}{4}ln\left|\dfrac{sinx-cosx+2}{sinx-cosx-2}\right|+C\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 2 2019
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 \(\int\dfrac{dt}{cos^3t}=32I\)
\(\Rightarrow32I=\dfrac{sint}{1-sin^2t}-32I+\int\dfrac{cost.dt}{cos^2t}\)
\(\Rightarrow64I=\dfrac{sint}{1-sin^2t}-\int\dfrac{d\left(sint\right)}{sin^2t-1}=\dfrac{sint}{1-sin^2t}-\dfrac{1}{2}ln\left|\dfrac{sint-1}{sint+1}\right|+C\)
\(\Rightarrow I=\dfrac{1}{64}\left(\dfrac{2x^4}{4-x^8}-\dfrac{1}{2}ln\left|\dfrac{x^4-2}{x^4+2}\right|\right)+C\)
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 11 2021
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. \(\int\dfrac{dx}{\sqrt{1-x^2}}=arcsin\left(x\right)+C\)
KR
31 tháng 1 2023
\(A=\int \frac{x}{\sqrt{x+2}}dx \\ = \int \frac{x+2-2}{\sqrt{x+2}}dx \\ = \int \sqrt{x+2}-2\frac{1}{\sqrt{x+2}}dx \\ = \frac{2}{3}(x+2)^{\frac{3}{2}}-4\sqrt{x+2}+C\)
\(B=\int \frac{sinx+cosx}{\sqrt[3]{1-sin2x}}dx \\ x=\frac{\pi}{4}-u, dx=-du \\ =- \int \frac{sin(\frac{\pi}{4}-u)+cos(\frac{\pi}{4}-u)}{\sqrt[3]{1-sin(\frac{\pi}{2}-2u)}}du \\ = - \int \frac{\frac{1}{\sqrt2}cosu+\frac{1}{\sqrt2}sinu+\frac{1}{\sqrt2}cosu-\frac{1}{\sqrt2}sinu}{\sqrt[3]{1-cos2u}}du \\ = -\int \frac{\frac{2}{\sqrt2}cosu}{\sqrt[3]{1-cos2u}}du \\ = -\sqrt2 \int \frac{cosu}{\sqrt[3]{1-cos^2u+sin^2u}}du \\ = -\sqrt2 \int \frac{cosu}{\sqrt[3]{2sin^2u}}du \\ v=sinu, dv=cosudu \\ = -\sqrt2 \int \frac{1}{\sqrt[3]{2v^2}}dv \\ = -\frac{\sqrt2}{\sqrt[3]2} \int v^{-\frac{2}{3}}dv \\ = -\frac{\sqrt2}{\sqrt[3]2} 3v^\frac{1}{3}+C \\ = -\frac{\sqrt2}{\sqrt[3]2} 3\sqrt[3]{sin(\frac{\pi}{4}-x)}+C \)
Câu 1:
\(A=\int \frac{2\sin x+\cos x}{3\sin x+2\cos x}dx\)
\(A=\int \frac{\frac{8}{13}(3\sin x+2\cos x)-\frac{1}{13}(3\cos x-2\sin x)}{3\sin x+2\cos x}dx\)
\(A=\frac{8}{13}\int dx-\frac{1}{13}\int \frac{(3\cos x-2\sin x)dx}{3\sin x+2\cos x}\)
\(A=\frac{8}{13}x-\frac{1}{13}\int \frac{d(3\sin x+2\cos x)}{3\sin x+2\cos x}\)
\(A=\frac{8}{13}x-\frac{1}{13}\ln |3\sin x+2\cos x|+c\)
Câu 2:
Ta có: \(I=\int \frac{x^3}{x^4+3x^2+2}dx=\int \frac{x^3}{(x^2+1)(x^2+2)}dx\)
\(=\int x^3\left(\frac{1}{x^2+1}-\frac{1}{x^2+2}\right)dx=\int \frac{x^3dx}{x^2+1}-\int \frac{x^3}{x^2+2}dx\)
\(=\frac{1}{2}\int \frac{x^2d(x^2+1)}{x^2+1}-\frac{1}{2}\int \frac{x^2d(x^2+2)}{x^2+2}\)
\(=\frac{1}{2}\int \left(1-\frac{1}{x^2+1}\right)d(x^2+1)-\frac{1}{2}\int \left(1-\frac{2}{x^2+2}\right)d(x^2+2)\)
\(=\frac{1}{2}\int d(x^2+1)-\frac{1}{2}\int \frac{d(x^2+1)}{x^2+1}-\frac{1}{2}\int d(x^2+2)+\int \frac{d(x^2+2)}{x^2+2}\)
\(=\frac{x^2+1}{2}-\frac{1}{2}\ln |x^2+1|-\frac{x^2+2}{2}+\ln |x^2+2|+c\)
\(=\ln |x^2+2|-\frac{1}{2}\ln |x^2+1|+c\)