1) Đặt \(2+lnx=t\Leftrightarrow x=e^{t-2}\Rightarrow dx=e^{t-2}dt\)

\(I_1=\int\left(\frac{t-2}{t}\right)^2\cdot e^{t-2}\cdot dt=\int\left(1-\frac{4}{t}+\frac{4}{t^2}\right)e^{t-2}dt\\ =\int e^{t-2}dt-4\int\frac{e^{t-2}}{t}dt+4\int\frac{e^{t-2}}{t^2}dt\)

Có:

\(4\int\frac{e^{t-2}}{t^2}dt=-4\int e^{t-2}\cdot d\left(\frac{1}{t}\right)=-\frac{4\cdot e^{t-2}}{t}+4\int\frac{e^{t-2}}{t}dt\\ \Leftrightarrow4\int\frac{e^{t-2}}{t^2}dt-4\int\frac{e^{t-2}}{t^{ }}dt=-\frac{4\cdot e^{t-2}}{t}\)

Vậy \(I_1=\int e^{t-2}dt-\frac{4\cdot e^{t-2}}{t}=e^{t-2}-\frac{4e^{t-2}}{t}+C\)

3) Đặt \(t=\sqrt{1+\sqrt[3]{x^2}}\Rightarrow t^2-1=\sqrt[3]{x^2}\Leftrightarrow x^2=\left(t^2-1\right)^3\)

\(d\left(x^2\right)=d\left[\left(t^2-1\right)^3\right]\Leftrightarrow2x\cdot dx=6t\left(t^2-1\right)^2\cdot dt\)

\(I_3=\int\frac{3t\left(t^2-1\right)^2}{t}dt=3\int\left(t^4-2t^2+1\right)dt=...\)

5) Đặt \(\frac{2+x}{2-x}=4t^3\Leftrightarrow4t^3=\frac{4}{2-x}-1\)

\(d\left(4t^3\right)=d\left(\frac{4}{2-x}-1\right)\Leftrightarrow3t^2dt=\frac{1}{\left(2-x\right)^2}dx\)

\(I_5=\int\frac{3t^2}{t\sqrt[3]{4}}dt=\frac{3}{\sqrt[3]{4}}\int tdt=...\)

Lời giải:

Đặt \(x=2\sin t( \frac{-\pi}{2}\leq t\leq \frac{\pi}{2})\)

Khi đó \(A=\int^{\frac{3}{2}}_{0}\frac{dx}{\sqrt{(4-x^2)^9}}=\int ^{\sin ^-1\left(\frac{3}{4}\right)}_{0}\frac{d(2\sin t)}{\sqrt{(4-4\sin^2 x)^9}}=\frac{1}{2^8}\int ^{\sin ^-1\left(\frac{3}{4}\right)}_{0}\frac{dt}{\cos^8 x}\)

Xét \(\int ^{\sin ^-1\left(\frac{3}{4}\right)}_{0}\frac{dt}{\cos^8 x}=\int ^{\sin ^-1\left(\frac{3}{4}\right)}_{0}\frac{d(\tan x)}{\cos ^6x}=\int ^{\sin ^-1\left(\frac{3}{4}\right)}_{0}\frac{(\sin^2x+\cos^2x)^3d(\tan x)}{\cos^6 x}\)

\(=\int ^{\sin ^-1\left(\frac{3}{4}\right)}_{0}(\tan^2 x+1)^3d(\tan x)=\int ^{\sin ^-1\left(\frac{3}{4}\right)}_{0}(\tan^6x+1+3\tan ^4x+3\tan ^2x)d(\tan x)\)

\(=\left.\begin{matrix} \sin^{-1}\left(\frac{3}{4}\right)\\ 0\end{matrix}\right|\left ( \frac{\tan ^7x}{7}+\tan x+\frac{3\tan^5x}{5}+\tan^3x \right )\)

\(\Rightarrow A=\left.\begin{matrix} \sin^{-1}\left(\frac{3}{4}\right)\\ 0\end{matrix}\right|\left ( \frac{\tan ^7x}{7}+\tan x+\frac{3\tan^5x}{5}+\tan^3x \right ).\frac{1}{2^8}\approx 0,015862\)

P/s: Kiểm tra kết quả tại http://www.wolframalpha.com/input/?i=integral+of+%5Csqrt%7B(4-x%5E2)%5E%7B-9%7D%7D+from+0+to+%5Cfrac%7B3%7D%7B2%7D

1/ \(\int\limits^e_1\left(x+\frac{1}{x}+\frac{1}{x^2}\right)dx=\left(\frac{x^2}{2}+lnx-\frac{1}{x}\right)|^e_1=\frac{e^2}{2}-\frac{1}{e}+\frac{3}{2}\)

2/ \(\int\limits^2_1\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)dx=\int\limits^2_1\left(x\sqrt{x}+1\right)dx=\int\limits^2_1\left(x^{\frac{3}{2}}+1\right)dx\)

\(=\left(\frac{2}{5}.x^{\frac{5}{2}}+x\right)|^2_1=\frac{8\sqrt{2}-7}{5}\)

3/

\(\int\limits^2_1\frac{2x^3-4x+5}{x}dx=\int\limits^2_1\left(2x^2-4+\frac{5}{x}\right)dx=\left(\frac{2}{3}x^3-4x+5lnx\right)|^2_1=\frac{2}{3}+5ln2\)

4/ \(\int\limits^2_1x^2\left(3x-1\right)\frac{2}{x}dx=\int\limits^2_1\left(6x^2-2x\right)dx=\left(2x^3-x^2\right)|^2_1=11\)

\(\int\limits^1_{\sqrt{ }3}\)\(\sqrt{\left(1+x^2\right)}\)\(dx\)

tks cậu

câu trả lời của mình cũng giống Hương Trà

Câu 1,2,3 Ez quá rồi :3

Câu 4:

Tổng quát:

\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}.\) Game là dễ :v

Câu 5 ko khác câu 4 lắm :v

Câu 5: 

Tổng quát:

\(\frac{1}{\sqrt{a}-\sqrt{a+1}}=\frac{\sqrt{a}+\sqrt{a+1}}{a-a-1}=-\sqrt{a}-\sqrt{a+1}.\) Game là dễ :v

Sao làm hổng ai bảo đú.n/g vậy :(((

\(=\int\left(6x^2-\dfrac{4}{x}+sin3x-cos4x+e^{2x+1}+9^{x-1}+\dfrac{1}{cos^2x}-\dfrac{1}{sin^2x}\right)dx\)

\(=2x^3-4ln\left|x\right|-\dfrac{1}{3}cos3x-\dfrac{1}{4}sin4x+\dfrac{1}{2}e^{2x+1}+\dfrac{9^{x-1}}{ln9}+tanx+cotx+C\)