\(\int^{44}_{4^4\sqrt[4]{4}}\) =
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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. \(\int\dfrac{dx}{\sqrt{1-x^2}}=arcsin\left(x\right)+C\)
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 \(\int\dfrac{dt}{cos^3t}=32I\)
\(\Rightarrow32I=\dfrac{sint}{1-sin^2t}-32I+\int\dfrac{cost.dt}{cos^2t}\)
\(\Rightarrow64I=\dfrac{sint}{1-sin^2t}-\int\dfrac{d\left(sint\right)}{sin^2t-1}=\dfrac{sint}{1-sin^2t}-\dfrac{1}{2}ln\left|\dfrac{sint-1}{sint+1}\right|+C\)
\(\Rightarrow I=\dfrac{1}{64}\left(\dfrac{2x^4}{4-x^8}-\dfrac{1}{2}ln\left|\dfrac{x^4-2}{x^4+2}\right|\right)+C\)
TN
1
AH
Akai Haruma
Giáo viên
28 tháng 12 2016
Câu 1:Gọi biểu thức là $A$. Đặt \(\sqrt{e^x-1}=t\)
\(\Rightarrow e^x=t^2+1\Rightarrow d(e^x)=d(t^2+1)=2tdt=e^xdx=(t^2+1)dx\)
\(\Rightarrow \int \frac{2t^2}{t^2+1}dt=\int \left (2-\frac{2}{t^2+1} \right)dt\)
Đặt \(t=\tan m\Rightarrow dt=\frac{dm}{\cos^2 m}\Rightarrow \int \frac{2dt}{t^2+1}=\int 2dm=2m\)
\(\Rightarrow A=2t-2m+c=2\sqrt{e^x-1}-2\tan ^{-1} (\sqrt{e^x-1})+c\)
Câu 2: Đặt \(x=\tan t\Rightarrow dx=\frac{dt}{\cos^2 t}, x^2+1=\frac{1}{\cos^2 t}\) với \(\frac{-\pi}{2} < t< \frac{\pi}{2}\)
Gọi biểu thức là $B$. Ta có
\(B=\int \frac{\cos t dt}{\sin ^4t}=\int \frac{d(\sin t)}{\sin^4 t}=\frac{-\sin ^{-3} t}{3}+c\) \(=-\frac{\sqrt{(x^2+1)^3}}{3x^3}+c\)
AH
Akai Haruma
Giáo viên
5 tháng 2 2018
Lời giải:
Ta có : \(P=\int x^7\sqrt{x^4+4}dx=\int x^4\sqrt{x^4+4}(x^3dx)\)
\(=\int x^4\sqrt{x^4+4}\frac{d(x^4+4)}{4}=\frac{1}{4}\int x^4\sqrt{x^4+4}d(x^4+4)\)
Đặt \(\sqrt{x^4+4}=t\Rightarrow x^4=t^2-4\)
Khi đó:
\(P=\frac{1}{4}\int (t^2-4)td(t^2)=\frac{1}{2}\int (t^2-4)t^2dt\)
\(=\frac{1}{2}\int t^4dt-2\int t^2dt=\frac{t^5}{10}-\frac{2t^3}{3}+c\)
\(=\frac{1}{10}\sqrt{(x^4+4)^5}-\frac{2}{3}\sqrt{(x^4+4)^3}+c\)
Vậy \(m=\frac{1}{10}; n=\frac{2}{3}\Rightarrow mn=\frac{1}{15}\)