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1/ \(4\int\limits^5_{-3}f'\left(x\right)dx=4f\left(x\right)|^5_{-3}=4\left[f\left(5\right)-f\left(-3\right)\right]=4.\left(9-1\right)=32\)
2/ \(\int\left(2x+1\right)e^xdx\)
\(\left\{{}\begin{matrix}u=2x+1\\dv=e^xdx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=2dx\\v=e^x\end{matrix}\right.\)
\(\Rightarrow\int\left(2x+1\right)e^xdx=\left(2x+1\right)e^x-2\int e^xdx=\left(2x+1\right)e^x-2e^x\)
P/s: Bạn tự thay cận vô nhé!
Note: \(\sqrt{\dfrac{1}{4x}+\dfrac{\sqrt{x}+e^x}{\sqrt{x}.e^{2x}}}=\sqrt{\dfrac{1}{4x}+\dfrac{1}{e^x.\sqrt{x}}+\dfrac{1}{e^{2x}}}=\sqrt{\left(\dfrac{1}{2\sqrt{x}}+\dfrac{1}{e^x}\right)^2}=\dfrac{1}{2\sqrt{x}}+\dfrac{1}{e^x}\)
Vấn đề bây giờ có lẽ đã quá đơn giản
\(\int\limits^1_0\dfrac{xdx}{\left(x+2\right)^2}=\int\limits^1_0\dfrac{1}{x+2}dx-\int\limits^1_0\dfrac{2}{\left(x+2\right)^2}dx=ln\left(x+2\right)|^1_0+\dfrac{2}{x+2}|^1_0=ln3-ln2-\dfrac{1}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{3}\\b=-1\\c=1\end{matrix}\right.\)
\(\int\limits^2_1\dfrac{lnx}{x^2}.dx\)
Dat \(\left\{{}\begin{matrix}u=lnx\\dv=\dfrac{dx}{x^2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=-\dfrac{1}{x}\end{matrix}\right.\)
\(\Rightarrow\int\limits^2_1\dfrac{lnx}{x^2}dx=lnx.\left(-\dfrac{1}{x}\right)|^2_1+\int\limits^2_1\dfrac{1}{x^2}.dx\)
\(=lnx.\left(-\dfrac{1}{x}\right)|^2_1+\left(-\dfrac{1}{x}\right)|^2_1=\left(-\dfrac{1}{2}\right).ln2+ln1-\dfrac{1}{2}+1\)
\(=\dfrac{1}{2}-\dfrac{1}{2}ln2\Rightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{2}\\b=1\\c=2\end{matrix}\right.\Rightarrow P=2a+3b+c=-1+3+2=4\)
Chọn B
Đặt u = ln ( x + 1 ) d v = d x
⇒ d u = 1 x + 1 d x v = x + 1
Khi đó
∫ 1 2 ln ( x + 1 ) d x = ( x + 1 ) ln ( x + 1 ) 1 2 - ∫ 1 2 d x = 3 ln 3 - 2 ln 2 - 1
Vậy a=3; b=-2; c=-1 ⇒ S = a + b + c = 0
