Giải giúp mình câu tích phân với ạ
\(\int\limits^1_0x^2e^{x^3}+\frac{\sqrt[4]{x}}{1+\sqrt{x}}dx\)
Tính các tích phân sau
1.I=\(\int\limits^{\frac{\Pi}{4}}_0\) (x+1)sin2xdx
2.I=\(\int\limits^2_1\frac{x^2+3x+1}{x^2+x}dx\)
3.I=\(\int\limits^2_1\frac{x^2-1}{x^2}lnxdx\)
4. I=\(\int\limits^1_0x\sqrt{2-x^2}dx\)
5.I=\(\int\limits^1_0\frac{\left(x+1\right)^2}{x^2+1}dx\)
6. I=\(\int\limits^5_1\frac{dx}{1+\sqrt{2x-1}}\)
7. I=\(\int\limits^3_1\frac{1+ln\left(x+1\right)}{x^2}dx\)
8.I=\(\int\limits^1_0\frac{x^3}{x^4+3x^2+2}dx\)
9. I=\(\int\limits^{\frac{\Pi}{4}}_0x\left(1+sin2x\right)dx\)
10. I=\(\int\limits^3_0\frac{x}{\sqrt{x+1}}dx\)
thầy giúp em câu tích phân này với ạ
\(\int\limits^2_0\left(x-2\right)\left(\sqrt{\frac{x}{4-x}}\right)dx\)
Tính tích phân m.n giúp e với ạ ///
\(\int\limits^{2\sqrt{3}}_2\frac{\sqrt{3}}{x\sqrt{x^2-3}}dx\)
đặt \(x=\frac{\sqrt{3}}{cost};\forall t\in\left(0;\frac{\pi}{2}\right)\Rightarrow tant>0\)
\(dx=d\left(\frac{\sqrt{3}}{cost}\right)=\frac{-\sqrt{3}sint}{cos^2t}dt\)
Thay vào, ta có \(\int\frac{\sqrt{3}\cdot\frac{-\sqrt{3}sint}{cos^2t}}{\frac{\sqrt{3}}{cost}\sqrt{\frac{3}{cos^2t}-3}}dt=\int\frac{-3\cdot\frac{sint}{cos^2t}}{\frac{3}{cost}\cdot\sqrt{tan^2t}}dt=\int\frac{-sint}{cost\cdot tant}dt=-\int dt=-t+C\)
Bây giờ thay t vào là ra
Tính :
a) \(\int\limits^3_0\dfrac{x}{\sqrt{1+x}}dx\)
b) \(\int\limits^{64}_1\dfrac{1+\sqrt{x}}{\sqrt[3]{x}}dx\)
c) \(\int\limits^2_0x^2e^{3x}dx\)
d) \(\int\limits^{\pi}_0\sqrt{1+\sin2x}dx\)
Tính tích phân : \(I=\int\limits^2_0\frac{x^5}{\sqrt{x^3+1}}dx\)
Ta có :\(I=\int\limits^2_0\frac{x^2x^3}{\sqrt{x^3+1}}dx\)
Đặt \(t=\sqrt{x^3+1}\) khi đó với x=0 thì t=1,x=2 thì t=3
và \(dt=\frac{3x^2}{2\sqrt{x^3+1}}dx\Rightarrow\frac{x^2}{\sqrt{x^3+1}}dx=\frac{2}{3}dt,x^3=t^2-1\)
Suy ra \(I=\frac{2}{3}\int\limits^3_1\left(t^2-1\right)dt=\frac{2}{3}\left(\frac{1}{3}t^2-t\right)|^3_1=\frac{2}{3}\left(\frac{26}{3}-2\right)=\frac{40}{9}\)
Vậy \(I=\int\limits^2_0\frac{x^5}{\sqrt{x^3+1}}dx=\frac{40}{9}\)
Tính các tích phân sau :
a) \(\int\limits^4_{-2}\left(\dfrac{x-2}{x+3}\right)^2dx\) (đặt \(t=x+3\) )
b) \(\int\limits^6_{-4}\left|x+3\right|-\left|x-4\right|dx\)
c) \(\int\limits^2_{-3}\dfrac{dx}{\sqrt{x+7}+3}\) (đặt \(t=\sqrt{x+7}\) hoặc \(t=\sqrt{x+7}+3\) )
d) \(\int\limits^{\dfrac{\pi}{2}}_0\dfrac{\cos x}{1+4\sin x}dx\)
e) \(\int\limits^2_1\dfrac{x^9}{x^{10}+4x^5+4}dx\) (đặt \(t=x^5\) )
g) \(\int\limits^3_0\left(x+2\right)e^{2x}dx\)
h) \(\int\limits^5_2\dfrac{\sqrt{4+x}}{x}dx\) (đặt \(t=\sqrt{4+x}\) )
Tính tích phân bằng định nghĩa và các tính chất:
1. \(\int\limits^e_1\left(x+\frac{1}{x}+\frac{1}{x^2}\right)dx\)
2. \(\int\limits^2_1\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)dx\)
3. \(\int\limits^2_1\frac{2x^3-4x+5}{x}dx\)
4. \(\int\limits^2_1x^2\left(3x-1\right)\frac{2}{x}dx\)
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/ \(\int\limits^2_1x^2\left(3x-1\right)\frac{2}{x}dx=\int\limits^2_1\left(6x^2-2x\right)dx=\left(2x^3-x^2\right)|^2_1=11\)
\(\int\limits^{\frac{\pi}{3}}_0\frac{sinx}{cosx\sqrt{3+sin^2x}}dx\)
\(\int\limits^{ln8}_0\frac{e^x}{1+\sqrt{3e^x+1}}dx\)
Tính tích phân sau :
\(\int\limits^1_0x\ln\left(1+x^2\right)dx\)
\(=\frac{1}{2}\int\limits^1_0\ln\left(1+x^2\right)d\left(1+x^2\right)=\frac{1}{2}\left[\left(1+x^2\right)\ln\left(1+x^2\right)\right]|^1_0-\int\limits^1_0d\left(1+x^2\right)\)
\(=\frac{1}{2}\left[2\ln2-\left(1+x^2\right)|^1_0\right]=\frac{\left(2\ln2-1\right)}{2}\)