\(\int\left(3x^2-2x-4\right)dx=x^3-x^2-4x+C\)

\(\int\left(sin3x-cos4x\right)dx=-\dfrac{1}{3}cos3x-\dfrac{1}{4}sin4x+C\)

\(\int\left(e^{-3x}-4^x\right)dx=-\dfrac{1}{3}e^{-3x}-\dfrac{4^x}{ln4}+C\)

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 \(\left\{{}\begin{matrix}u=lnx\\dv=xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=\dfrac{1}{2}x^2\end{matrix}\right.\)

\(\Rightarrow I=\dfrac{1}{2}x^2.lnx-\dfrac{1}{2}\int xdx=\dfrac{1}{2}x^2.lnx-\dfrac{1}{4}x^2+C\)

a/ \(\int\dfrac{x^2-3x+1}{x}dx=\int\left(x-3+\dfrac{1}{x}\right)dx=\int x.dx-3x+\int\dfrac{dx}{x}=\dfrac{1}{2}.x^2-3x+ln\left|x\right|+C\)

b/ \(I=\int x.e^{2x}dx\)

\(\left\{{}\begin{matrix}u=x\\dv=e^{2x}dx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=dx\\v=\dfrac{1}{2}e^{2x}\end{matrix}\right.\)

\(\Rightarrow I=\dfrac{1}{2}.x.e^{2x}-\dfrac{1}{2}\int e^{2x}.dx=\dfrac{1}{2}x.e^{2x}-\dfrac{1}{4}e^{2x}\)

Chọn A

1: \(\int \sin\left(\frac{\pi}{4} - x\right) dx\)

mà ta có công thức: \(\int \sin(ax + b) dx = -\frac{1}{a}\cos(ax + b) + C\) nên với a=-1; \(b=\frac{\pi}{4}\) thì

\(\int\sin\left(\frac{\pi}{4}-x\right)dx=-\frac{1}{-1}\cos\left(\frac{\pi}{4}-x\right)+C=\cos\left(\frac{\pi}{4}-x\right)+C\)

2: \(\int \frac{7}{\cos^2(3-x)} dx = \frac{7}{-1} \tan(3-x) = -7\tan(3-x)\)

\(\int 8\sin(9-3x) dx = 8 \cdot \left(-\frac{1}{-3}\right)\cos(9-3x) = \frac{8}{3}\cos(9-3x)\)

\(\int -\frac{1}{x} dx = -\ln\vert{}x\vert{}\)

\(\int \frac{6}{3-2x} dx = 6 \cdot \left(-\frac{1}{2}\right)\ln\vert{}3-2x\vert{} = -3\ln\vert{}3-2x\vert{}\)

\(\int \sqrt{x} dx = \int x^{\frac{1}{2}} dx = \frac{x^{\frac{3}{2}}}{\frac{3}{2}} = \frac{2}{3}x\sqrt{x}\)

\(\int \left( \frac{7}{\cos^2(3-x)} + 8\sin(9-3x) - \frac{1}{x} + \frac{6}{3-2x} + \sqrt{x} \right) dx\)

\(= -7\tan(3-x) + \frac{8}{3}\cos(9-3x) - \ln\vert{}x\vert{} - 3\ln\vert{}3-2x\vert{} + \frac{2}{3}x\sqrt{x} + C\)

3: \(\int \frac{7}{\cos^2 x} dx = 7\tan x\)

\(\int -\frac{8}{2x+1} dx = -8 \cdot \frac{1}{2} \ln\vert{}2x+1\vert{} = -4\ln\vert{}2x+1\vert{}\)

\(\int 9^{2x+1} dx = \frac{1}{2} \cdot \frac{9^{2x+1}}{\ln 9} = \frac{9^{2x+1}}{4\ln 3}\)

\(\int e^{5-2x} dx = -\frac{1}{2}e^{5-2x}\)

\(\int 8 dx = 8x\)

\(\int \left( \frac{7}{\cos^2 x} - \frac{8}{2x+1} + 9^{2x+1} + e^{5-2x} + 8 \right) dx\)

\(= 7\tan x - 4\ln\vert{}2x+1\vert{} + \frac{9^{2x+1}}{4\ln 3} - \frac{1}{2}e^{5-2x} + 8x + C\)

4: \(\int \frac{4}{x} dx = 4\ln\vert{}x\vert{}\)

\(\int -x^{-\frac{1}{2}} dx = -\frac{x^{\frac{1}{2}}}{\frac{1}{2}} = -2\sqrt{x}\)

\(\int 5x^4 dx = 5 \cdot \frac{x^5}{5} = x^5\)

\(\int -6x^6 dx = -6 \cdot \frac{x^7}{7} = -\frac{6}{7}x^7\)

\(\int\frac{3 - \sqrt{x} + 5x^5 - 6x^7 + 1}{x}dx\)

\(=\int\frac{4 - x^{\frac{1}{2}} + 5x^5 - 6x^7}{x}dx\)

\(= \int \left( \frac{4}{x} - x^{-\frac{1}{2}} + 5x^4 - 6x^6 \right) dx\)

\(= 4\ln\vert{}x\vert{} - 2\sqrt{x} + x^5 - \frac{6}{7}x^7 + C\)

\(\int\dfrac{xe^x}{\left(x+1\right)^2}dx\)

\(=\int e^x.\dfrac{\left(x+1\right)-1}{\left(x+1\right)^2}dx=\int e^x.[\dfrac{1}{x+1}-\dfrac{1}{\left(x+1\right)^2}]dx\)

\(=\int\dfrac{e^x}{x+1}dx-\int\dfrac{e^x}{\left(x+1\right)^2}dx=\dfrac{1}{x+1}e^x+\int\dfrac{e^x}{\left(x+1\right)^2}dx-\int\dfrac{e^x}{\left(x+1\right)^2}dx\)

\(=\dfrac{e^x}{x+1}+C\)

Ko chac :v

\(I=\int\dfrac{x.e^x}{\left(x+1\right)^2}dx\)

Đặt \(\left\{{}\begin{matrix}u=xe^x\\dv=\dfrac{1}{\left(x+1\right)^2}dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=e^x\left(x+1\right)dx\\v=-\dfrac{1}{x+1}\end{matrix}\right.\)

\(I=\dfrac{-xe^x}{x+1}+\int e^xdx=\dfrac{-xe^x}{x+1}+e^x+C=\dfrac{e^x}{x+1}+C\)

- 1 2 e - x 2 + C

@Nguyễn Việt Lâm

Đây là 1 tích phân không tính được bạn Thơ ạ

Đáp án B

Ta có  y ' = e − x − x 2 e − x ⇒ e − x − x e − x > 0 ⇔ 1 − x > 0 ⇔ x < 1