Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1. Đặt \(\left\{{}\begin{matrix}u=x\\dv=\dfrac{dx}{sin^2x}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=dx\\v=-cotx\end{matrix}\right.\)
Do đó I= \(-x.cotx+\int cotxdx\)= \(-xcotx+ln\left|sinx\right|\)
2. Đặt \(\left\{{}\begin{matrix}u=x+1\\dv=\dfrac{dx}{e^x}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=dx\\v=-e^{-x}\end{matrix}\right.\)
Do đó I= \(-\left(x+1\right)e^{-x}+\int e^{-x}dx\)=\(-\left(x+1\right)e^{-x}-e^{-x}\)
=\(-\left(x+2\right)e^{-x}\)
\(\int\left(\frac{1}{x}-2x\right)dx=ln\left|x\right|-x^2+C\)
\(\int cos2xdx=\frac{1}{2}sin2x+C\)
\(\int\frac{1}{x^2-4x+4}dx=\int\frac{d\left(x-2\right)}{\left(x-2\right)^2}=-\frac{1}{\left(x-2\right)}+C=\frac{1}{2-x}+C\)
\(\int\limits^4_1\frac{1}{2\sqrt{x}}dx=\sqrt{x}|^4_1=\sqrt{4}-\sqrt{1}=1\)
\(I=\int\limits^1_0\left(2x+1\right)e^xdx\)
Đặt \(\left\{{}\begin{matrix}u=2x+1\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=2dx\\v=e^x\end{matrix}\right.\)
\(\Rightarrow I=\left(2x+1\right)e^x|^1_0-\int\limits^1_02e^xdx=3e-1-2e^x|^1_0=e+3\)
\(I=\int\limits^1_0\frac{xdx}{\sqrt{3x+1}+\sqrt{2x+1}}=\int\limits^1_0\frac{x\left(\sqrt{3x+1}-\sqrt{2x+1}\right)}{x}dx\)
\(=\int\limits^1_0\left(\left(3x+1\right)^{\frac{1}{2}}-\left(2x+1\right)^{\frac{1}{2}}\right)dx=\left[\frac{2}{9}\left(3x+1\right)^{\frac{3}{2}}-\frac{1}{3}\left(2x+1\right)^{\frac{3}{2}}\right]|^1_0\)
\(=\frac{2}{9}\sqrt{4^3}-\frac{1}{3}\sqrt{3^3}-\frac{2}{9}+\frac{1}{3}=\frac{17-9\sqrt{3}}{9}\)
\(\Rightarrow a+b=17-9=8\)
i3 = i2 .i = -i; i4 = i2 .i2 = (-1)(-1) = 1; i5 = i4 .i = i
Nếu n = 4q + r, 0 ≤ r < 4 thì
1) in = ir = i nếu r = 1
2) in = ir = -1 nếu r = 2
3) in = ir = -i nếu r = 3
4) in = ir = 1 nếu r = 4
i3 = i2 .i = -i; i4 = i2 .i2 = (-1)(-1) = 1; i5 = i4 .i = i
Nếu n = 4q + r, 0 ≤ r < 4 thì
1) in = ir = i nếu r = 1
2) in = ir = -1 nếu r = 2
3) in = ir = -i nếu r = 3
4) in = ir = 1 nếu r = 4
I 3 = 2 3 , I 5 = 8 15