Nhìn đề dữ dội y hệt cr của tui z :( Để làm từ từ 

Lập bảng xét dấu cho \(\left|x^2-1\right|\) trên đoạn \(\left[-2;2\right]\)

\(\left(-2;-1\right):+\)

\(\left(-1;1\right):-\)

\(\left(1;2\right):+\)

\(\Rightarrow I=\int\limits^{-1}_{-2}\left|x^2-1\right|dx+\int\limits^1_{-1}\left|x^2-1\right|dx+\int\limits^2_1\left|x^2-1\right|dx\)

\(=\int\limits^{-1}_{-2}\left(x^2-1\right)dx-\int\limits^1_{-1}\left(x^2-1\right)dx+\int\limits^2_1\left(x^2-1\right)dx\)

\(=\left(\dfrac{x^3}{3}-x\right)|^{-1}_{-2}-\left(\dfrac{x^3}{3}-x\right)|^1_{-1}+\left(\dfrac{x^3}{3}-x\right)|^2_1\)

Bạn tự thay cận vô tính nhé :), hiện mình ko cầm theo máy tính 

2/ \(I=\int\limits^e_1x^{\dfrac{1}{2}}.lnx.dx\)

\(\left\{{}\begin{matrix}u=lnx\\dv=x^{\dfrac{1}{2}}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=\dfrac{2}{3}.x^{\dfrac{3}{2}}\end{matrix}\right.\)

\(\Rightarrow I=\dfrac{2}{3}.x^{\dfrac{3}{2}}.lnx|^e_1-\dfrac{2}{3}\int\limits^e_1x^{\dfrac{1}{2}}.dx\)

\(=\dfrac{2}{3}.x^{\dfrac{3}{2}}.lnx|^e_1-\dfrac{2}{3}.\dfrac{2}{3}.x^{\dfrac{3}{2}}|^e_1=...\)

3/ \(I=\int\limits^{\dfrac{\pi}{2}}_0e^{\sin x}.\cos x.dx+\int\limits^{\dfrac{\pi}{2}}_0\cos^2x.dx\)

Xét \(A=\int\limits^{\dfrac{\pi}{2}}_0e^{\sin x}.\cos x.dx\)

\(t=\sin x\Rightarrow dt=\cos x.dx\Rightarrow A=\int\limits^{\dfrac{\pi}{2}}_0e^t.dt=e^{\sin x}|^{\dfrac{\pi}{2}}_0\)

Xét \(B=\int\limits^{\dfrac{\pi}{2}}_0\cos^2x.dx\)

\(=\int\limits^{\dfrac{\pi}{2}}_0\dfrac{1+\cos2x}{2}.dx=\dfrac{1}{2}.\int\limits^{\dfrac{\pi}{2}}_0dx+\dfrac{1}{2}\int\limits^{\dfrac{\pi}{2}}_0\cos2x.dx\)

\(=\dfrac{1}{2}x|^{\dfrac{\pi}{2}}_0+\dfrac{1}{2}.\dfrac{1}{2}\sin2x|^{\dfrac{\pi}{2}}_0\)

I=A+B=...

1)

\(I=\int\left(cos^2x-cos^2x\cdot sin^3x\right)dx\\ =\int cos^2x\cdot dx-\int cos^2x\cdot sin^3x\cdot dx\\ =\frac{1}{2}\int\left(cos2x+1\right)dx+\int cos^2x\left(1-cos^2x\right)d\left(cosx\right)\\ =\frac{1}{4}sin2x+\frac{1}{2}+\frac{cos^3x}{3}-\frac{cos^5x}{5}+C\)

....

2) Xét riêng mẫu số:

\(sin2x+2\left(1+sinx+cosx\right)\\ =\left(sin2x+1\right)+2\left(sinx+cosx\right)+1\\ =\left(sinx+cosx\right)^2+2\left(sinx+cosx\right)+1\\ =\left(sinx+cosx+1\right)^2\\ =\left[\sqrt{2}cos\left(x-\frac{\pi}{4}\right)+1\right]^2\)

Khi đó:

\(I_2=\int\frac{sin\left(x-\frac{\pi}{4}\right)}{\left[\sqrt{2}cos\left(x-\frac{\pi}{4}\right)+1\right]^2}dx\\ =-\frac{1}{\sqrt{2}}\int\frac{d\left[\sqrt{2}cos\left(x-\frac{\pi}{4}\right)+1\right]}{\left[\sqrt{2}cos\left(x-\frac{\pi}{4}\right)+1\right]^2}\\ =\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}cos\left(x-\frac{\pi}{4}\right)+1}+C=\frac{1}{2cos\left(x-\frac{\pi}{4}\right)+1}\)

...

Câu 1)

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

\(\int \left ( \frac{\ln}{x} \right )^2dx=\frac{-\ln^2x}{x}+2\int \frac{\ln x}{x^2}dx\)

Đặt \(\left\{\begin{matrix} t=\ln x\\ dk=\frac{1}{x^2}dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} dt=\frac{1}{x}dx\\ k=-\frac{1}{x}\end{matrix}\right.\Rightarrow \int \frac{\ln x}{x^2}dx=-\frac{\ln x}{x}+\int \frac{1}{x^2}dx=\frac{-\ln x}{x}-\frac{1}{x}\)

\(\Rightarrow I=\left.\begin{matrix} e\\ 1\end{matrix}\right|\left(\frac{-\ln^2 x}{x}-\frac{2\ln x}{x}-\frac{2}{x}\right)=2-\frac{5}{e}\)

Câu 2)

\(I=\int ^{\frac{\pi}{4}}_{0}\frac{x}{1+\cos 2x}dx=\frac{1}{2}\int ^{\frac{\pi}{4}}_{0}\frac{x}{\cos^2x}dx\)

Đặt \(\left\{\begin{matrix} u=x\\ dv=\frac{dx}{\cos^2x}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=\tan x\end{matrix}\right.\Rightarrow I=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|\frac{x\tan x}{2}-\frac{1}{2}\int^{\frac{\pi}{4}}_{0} \tan xdx\)

\(=\frac{\pi}{8}+\frac{1}{2}\int ^{\frac{\pi}{4}}_{0}\frac{d(\cos x)}{\cos x}=\frac{\pi}{8}+\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|\frac{\ln |\cos x|}{2}=\frac{\pi}{8}+\frac{\ln\frac{\sqrt{2}}{2}}{2}\)

Câu 3)

Đặt \(\left\{\begin{matrix} u=\ln (\cos x)\\ dv=\frac{dx}{\cos^2x}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{-\sin x}{\cos x}dx=-\tan xdx\\ v=\tan x\end{matrix}\right.\)

\(\Rightarrow I=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|\tan x\ln (\cos x)+\int ^{\frac{\pi}{4}}_{0}\tan^2xdx=\ln \frac{\sqrt{2}}{2}+\int ^{\frac{\pi}{4}}_{0}(\frac{1}{\cos^2x}-1)dx\)

\(=\ln\frac{\sqrt{2}}{2}+\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|(\tan x-x)=\ln \frac{\sqrt{2}}{2}-\frac{\pi}{4}+1\)

Đang học Lý mà thấy bài nguyên hàm hay hay nên nhảy vô luôn :b

\(I_1=\int\limits^1_0xf\left(x\right)dx\)

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

\(\Rightarrow\int xf\left(x\right)dx=\dfrac{1}{2}x^2f\left(x\right)-\dfrac{1}{2}\int x^2f'\left(x\right)dx\)

\(\Rightarrow\int\limits^1_0xf\left(x\right)dx=\dfrac{1}{2}x^2|^1_0-\dfrac{1}{2}\int\limits^1_0x^2f'\left(x\right)dx=\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{1}{2}\int\limits^1_0\left[f'\left(x\right)\right]^2dx=\dfrac{3}{10}\Rightarrow\int\limits^1_0x^2f'\left(x\right)dx=\dfrac{3}{5}\)

Đoạn này hơi rối xíu, ông để ý kỹ nhé, nhận thấy ta có 2 dữ kiện đã biết, là: \(\int\limits^1_0\left[f'\left(x\right)\right]^2dx=\dfrac{9}{5}and\int\limits^1_0x^2f'\left(x\right)dx=\dfrac{3}{5}\) có gì đó liên quan đến hằng đẳng thức, nên ta sẽ sử dụng luôn

\(\int\limits^1_0\left[f'\left(x\right)+tx^2\right]^2dx=0\)

\(\Leftrightarrow\int\limits^1_0\left[f'\left(x\right)\right]^2dx+2t\int\limits^1_0x^2f'\left(x\right)dx+t^2\int\limits^1_0x^4dx=0\)

\(\Leftrightarrow\dfrac{9}{5}+\dfrac{6}{5}t+\dfrac{1}{5}t^2=0\)  \(\left(\int\limits^1_0x^4dx=\dfrac{1}{5}x^5|^1_0=\dfrac{1}{5}\right)\)\(\)\(\Leftrightarrow t=-3\Rightarrow\int\limits^1_0\left[f'\left(x\right)-3x^2\right]^2dx=0\)

\(\Leftrightarrow f'\left(x\right)=3x^2\Leftrightarrow f\left(x\right)=x^3+C\)

\(\Rightarrow\int\limits^1_0f\left(x\right)dx=\int\limits^1_0x^3dx=\dfrac{1}{4}x^4|^1_0=\dfrac{1}{4}\)

P/s: Có gì ko hiểu hỏi mình nhé !

\(3\int\limits^1_0\left[f'\left(x\right).f^2\left(x\right)+\frac{1}{9}\right]dx\le2\int\limits^1_0\sqrt{f'\left(x\right)}f\left(x\right)dx\) (1)

Ta lại có:

\(3f'\left(x\right).f^2\left(x\right)+\frac{1}{3}\ge2\sqrt{f'\left(x\right)}.f\left(x\right)\)

\(\Rightarrow3\int\limits^1_0\left[f'\left(x\right).f^2\left(x\right)+\frac{1}{9}\right]\ge2\int\limits^1_0\sqrt{f'\left(x\right)}.f\left(x\right)dx\) (2)

Từ (1); (2) \(\Rightarrow3\int\limits^1_0\left[f'\left(x\right).f^2\left(x\right)+\frac{1}{9}\right]dx=2\int\limits^1_0\sqrt{f'\left(x\right)}.f\left(x\right)dx\)

Dấu "=" xảy ra khi và chỉ khi:

\(3f'\left(x\right).f^2\left(x\right)=\frac{1}{3}\Rightarrow3\int f'\left(x\right).f^2\left(x\right)dx=\int\frac{1}{3}dx\)

\(\Rightarrow f^3\left(x\right)=\frac{x}{3}+C\)

Thay \(x=0\Rightarrow f^3\left(0\right)=C\Rightarrow C=1\)

\(\Rightarrow f^3\left(x\right)=\frac{x}{3}+1\Rightarrow\int\limits^1_0f^3\left(x\right)dx=\int\limits^1_0\left(\frac{x}{3}+1\right)dx=\frac{7}{6}\)

\(I=\int\limits^1_0\frac{x^3+2x^2+3}{x+2}dx=\int\limits^1_0\left(x^2+\frac{3}{x+2}\right)dx=\left(\frac{x^3}{3}+3ln\left|x+2\right|\right)|^1_0\)

\(=\left(\frac{1}{3}+3ln3\right)-3ln2=\frac{1}{3}+3ln\frac{3}{2}\)

\(\Rightarrow a=b=3\Rightarrow S=18\)