Bài 2: Tích phân

\(I=\int\dfrac{\left(\sqrt{x+1}+\sqrt{x}\right)dx}{x+1-x}=\int\sqrt{x+1}dx+\int\sqrt{x}dx\)

Xet \(I_1=\int\sqrt{x+1}dx\)

\(t=x+1\Rightarrow dt=dx\Rightarrow I_1=\int\sqrt{t}.dt=\dfrac{2}{3}\left(x+1\right)^{\dfrac{3}{2}}+C\)

\(\Rightarrow I=\dfrac{2}{3}\left(x+1\right)^{\dfrac{3}{2}}+\dfrac{2}{3}x^{\dfrac{3}{2}}+C\)

P/s: Bạn tự thay cận vô ạ

\(I=\int\limits^{\dfrac{\pi}{4}}_0xsinxdx\)

Đặt \(\left\{{}\begin{matrix}u=x\\dv=sinxdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=dx\\v=-cosx\end{matrix}\right.\)

\(\Rightarrow I=-x.cosx|^{\dfrac{\pi}{4}}_0+\int\limits^{\dfrac{\pi}{4}}_0cosxdx=\left(-x.cosx+sinx\right)|^{\dfrac{\pi}{4}}_0=-\dfrac{\pi\sqrt{2}}{8}+\dfrac{\sqrt{2}}{2}\)

\(I_7=\int\limits^3_0x^2\sqrt{10-x^2}xdx\)

Đặt \(\sqrt{10-x^2}=t\Rightarrow x^2=10-t^2\Rightarrow xdx=tdt\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow t=\sqrt{10}\\x=3\Rightarrow t=1\end{matrix}\right.\)

\(I_7=\int\limits^1_{\sqrt{10}}\left(10-t^2\right)t.tdt=\int\limits^{\sqrt{10}}_1\left(t^4-10t^2\right)dt\)

\(=\left(\dfrac{1}{5}t^5-\dfrac{10}{3}t^3\right)|^{\sqrt{10}}_1\) (tới đây bạn tự tính ra kết quả nhé)

\(I_8=\int\limits^{\dfrac{\pi}{4}}_02sinx.cosx.cosxdx=-2\int\limits^{\dfrac{\pi}{4}}_0cos^2x.d\left(cosx\right)\)

\(=-\dfrac{2}{3}cos^3x|^{\dfrac{\pi}{4}}_0=...\)

\(I_9=\int\limits^{\dfrac{\pi}{2}}_0\dfrac{sinx}{1+3cosx}dx\)

Đặt \(u=cosx\Rightarrow du=-sinx.dx\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow u=1\\x=\dfrac{\pi}{2}\Rightarrow u=0\end{matrix}\right.\)

\(I_9=\int\limits^0_1\dfrac{-du}{1+3u}=\dfrac{1}{3}\int\limits^1_0\dfrac{d\left(3u+1\right)}{3u+1}=\dfrac{1}{3}ln\left(3u+1\right)|^1_0=\dfrac{1}{3}ln10\)

\(I_{10}=\int\limits^{\dfrac{\pi}{2}}_0sin^4x.\left(1-sin^2x\right)^2cosxdx\)

Đặt \(u=sinx\Rightarrow du=cosxdx\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow u=0\\x=\dfrac{\pi}{2}\Rightarrow u=1\end{matrix}\right.\)

\(I_{10}=\int\limits^1_0u^4\left(1-u^2\right)du=\int\limits^1_0\left(u^8-2u^6+u^4\right)du\)

\(=\left(\dfrac{1}{9}u^9-\dfrac{2}{7}u^7+\dfrac{1}{5}u^5\right)|^1_0=...\)

\(I_{11}=\int\limits^{\dfrac{\pi}{2}}_0\left(2x-1\right)cosxdx\)

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

\(\Rightarrow I_{11}=\left(2x-1\right)sinx|^{\dfrac{\pi}{2}}_0-2\int\limits^{\dfrac{\pi}{2}}_0sinxdx\)

\(=\pi-1+2cosx|^{\dfrac{\pi}{2}}_0=\pi-3\)

\(I_{12}=\int\limits^1_0\dfrac{e^xdx}{1+e^x}=\int\limits^1_0\dfrac{d\left(e^x+1\right)}{e^x+1}\)

\(=ln\left(e^x+1\right)|^1_0=ln\left(e+1\right)-ln2=ln\left(\dfrac{e+1}{2}\right)\)

Đặt \(\sqrt{mx}=u\Rightarrow x=\dfrac{u^2}{m}\Rightarrow dx=\dfrac{2udu}{m}\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow u=0\\x=m\Rightarrow u=\left|m\right|\end{matrix}\right.\)

\(I=\int\limits^{\left|m\right|}_0\left|\dfrac{u^4}{m^3}-u\right|.\dfrac{2u}{m}du\)

Xét hàm \(f\left(u\right)=\dfrac{u^4}{m^3}-u=\dfrac{u\left(u-m\right)\left(u^2+mu+m^2\right)}{m^3}\) với \(u\in\left(0;\left|m\right|\right)\)

Do \(u^2+mu+m^2>0\)

- Khi \(m< 0\Rightarrow u\left(u-m\right)>0\Rightarrow f\left(u\right)< 0\)

- Khi \(m>0\Rightarrow u\left(u-m\right)< 0\) ; \(\forall u\in\left(0;m\right)\Rightarrow f\left(u\right)< 0\)

\(\Rightarrow f\left(u\right)< 0\) ; \(\forall m\) và \(u\in\left(0;\left|m\right|\right)\)

\(\Rightarrow\left|f\left(u\right)\right|=-f\left(u\right)=u-\dfrac{u^4}{m^3}\)

\(\Rightarrow I=\int\limits^{\left|m\right|}_0\left(u-\dfrac{u^4}{m^3}\right)\dfrac{2u}{m}du=\int\limits^{\left|m\right|}_0\left(\dfrac{2}{m}u^2-\dfrac{2}{m^4}u^5\right)du\)

\(=\left(\dfrac{2}{3m}u^3-\dfrac{1}{3m^4}u^6\right)|^{\left|m\right|}_0=\dfrac{2\left|m^3\right|}{3m}-\dfrac{m^6}{3m^4}=3\)

\(\Leftrightarrow2m\left|m\right|-m^2=9\)

- Với \(m< 0\Rightarrow VT< 0\Rightarrow\) pt vô nghiệm

- Với \(m>0\Rightarrow m^2=9\Rightarrow m=3\)

1/ \(\left\{{}\begin{matrix}u=ln\left(x+1\right)\\dv=dx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=\dfrac{1}{x+1}dx\\v=x\end{matrix}\right.\)

\(\Rightarrow I=\int ln\left(1+x\right)dx=x.ln\left(x+1\right)-\int\dfrac{x}{x+1}dx\)

Xet \(I_1=\int\dfrac{x}{x+1}dx=\int(1-\dfrac{1}{x+1})dx=\int dx-\int\dfrac{dx}{x+1}=x-ln\left|x+1\right|\)

\(\Rightarrow I=x.ln\left(x+1\right)-x+ln\left|x+1\right|\)

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

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

\(I_1=\int\left(x-1\right)e^{2x+1}dx\)

\(\left\{{}\begin{matrix}t=x-1\\dy=e^{2x+1}dx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}dt=dx\\y=\dfrac{1}{2}.e^{2x+1}\end{matrix}\right.\)

\(\Rightarrow I_1=\dfrac{1}{2}\left(x-1\right)e^{2x+1}-\dfrac{1}{2}\int e^{2x+1}dx=\dfrac{1}{2}\left(x-1\right)e^{2x+1}-\dfrac{1}{4}e^{2x+1}\)

\(\Rightarrow I=\dfrac{1}{2}\left(x^2-2x\right)e^{2x+1}-\dfrac{1}{2}\left(x-1\right)e^{2x+1}+\dfrac{1}{4}e^{2x+1}\)

P/s: Bạn tự thay cận vô nhé!

1/ \(u=x-3\Rightarrow du=dx\Rightarrow I_3=\int\limits^4_0u^{\dfrac{1}{2}}du=\dfrac{2}{3}.u^{\dfrac{3}{2}}|^4_0=\dfrac{2}{3}.4^{\dfrac{2}{3}}=..\)

2/ \(K_3=\int\limits^1_0\dfrac{1}{\left(x-3\right)\left(x-2\right)}dx=\int\limits^1_0(\dfrac{1}{x-3}-\dfrac{1}{x-2})dx\)

\(=\int\limits^1_0\left(x-3\right)^{-1}dx-\int\limits^1_0\left(x-2\right)^{-1}dx\)

\(=\int\limits^1_0\left(x-3\right)^{-1}d\left(x-3\right)-\int\limits^1_0\left(x-2\right)^{-1}d\left(x-2\right)\)

\(=ln\left|x-3\right||^1_0-ln\left|x-2\right||^1_0=...\)

\(2I=\int\limits^4_0\left(e^x\sqrt{2x+1}+\dfrac{e^x}{\sqrt{2x+1}}\right)dx=\int\limits^4_0e^x\sqrt{2x+1}dx+\int\limits^4_0\dfrac{e^x}{\sqrt{2x+1}}dx=I_1+I_2\)

Xét \(I_1=\int\limits^4_0e^x\sqrt{2x+1}dx\)

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

\(\Rightarrow I_1=e^x.\sqrt{2x+1}|^4_0-\int\limits^4_0\dfrac{e^x}{\sqrt{2x+1}}dx=3e^4-1-I_2\)

Do đó:

\(2I=3e^4-1-I_2+I_2=3e^4-1\)

\(\Rightarrow I=\dfrac{3}{2}e^4-\dfrac{1}{2}\Rightarrow a=\dfrac{3}{2};b=4;c=-\dfrac{1}{2}\)

Bạn dùng gõ công thức để mọi người hiểu đề bài nhé ! 

\(I=\int\limits^{\sqrt{3}}_0\dfrac{x^2}{1+x^2}dx=\int\limits^{\sqrt{3}}_0\left(1-\dfrac{1}{1+x^2}\right)dx\)

\(=\left(x-arctan\left(x\right)\right)|^{\sqrt{3}}_0=\sqrt{3}-\dfrac{\pi}{3}\)

\(\dfrac{87552}{35}\)