Giải đáp bài tích phân: \[I=\int_{0}^{1}x^2\sqrt{x^2+1}dx\]

Đề bài: Tính tích phân sau: \[I=\int_{0}^{1}x^2\sqrt{x^2+1}dx\]

Hướng dẫn giải:

Cách 1: 

Đổi biến: \[x=tant, t\in \left ( -\frac{\pi}{2};\frac{\pi}{2} \right )\Rightarrow dx=\left (tan^2t+1 \right )dt\]

Đổi cận: \[x=0\Rightarrow t=0, x=1\Rightarrow t=\frac{\pi}{4}\]

Khi đó \[I=\int_{0}^{\frac{\pi}{4}}tan^2t.\left ( tan^2t+1 \right ).\sqrt{tan^2t+1}dt=\int_{0}^{\frac{\pi}{4}}\frac{sin^2t.cost}{cos^6t}dt\]

Đổi biến: \[u=sint, u\in\left [ -1;1 \right ] \Rightarrow du=costdt\]

Đổi cận: \[t=0\Rightarrow u=0,t=\frac{\pi}{4}\Rightarrow u=\frac{\sqrt{2}}{2}\]

Khi đó \[I=\int_{0}^{\frac{\sqrt{2}}{2}}\frac{u^2}{\left ( 1-u^2 \right )^3}du=\int_{0}^{\frac{\sqrt{2}}{2}}\left [\frac{1}{\left ( 1-u^2 \right )^3}-\frac{1}{\left ( 1-u^2 \right )^2} \right ]du=H-K\]

Tính K:

\[K=\int_{0}^{\frac{\sqrt{2}}{2}}\frac{1}{\left ( 1-u^2 \right )^2}du=\frac{1}{4}\int_{0}^{\frac{\sqrt{2}}{2}}\left [ \left (\frac{1}{1-u}+\frac{1}{1+u} \right )^2 \right ]du\]

\[=\frac{1}{4}\int_{0}^{\frac{\sqrt{2}}{2}}\left [ \frac{1}{\left ( 1-u \right )^2}+\frac{1}{\left (1+u \right )^2}+\frac{1}{1-u}+\frac{1}{1+u} \right ]du\]

\[=\left.\begin{matrix} \frac{1}{4}\left ( \frac{1}{1-u}-\frac{1}{1+u}+ln\left | \frac{1+u}{1-u} \right | \right ) \end{matrix}\right|_{0}^{\frac{\sqrt{2}}{2}}=\frac{\sqrt{2}+ln\left (\sqrt{2} +1 \right )}{2}\]

Tính H:

\[H=\int_{0}^{\frac{\sqrt{2}}{2}}\frac{du}{\left ( 1-u^2 \right )^3}\]

\[=\frac{1}{8}\int_{0}^{\frac{\sqrt{2}}{2}}\left [ \frac{1}{\left ( 1-u \right )^3}+\frac{1}{\left ( 1+u \right )^3}+\frac{3}{\left ( 1-u \right )\left ( 1+u \right )}\left ( \frac{1}{1-u}+\frac{1}{1+u} \right ) \right ]du\]

\[=\frac{1}{8}\int_{0}^{\frac{\sqrt{2}}{2}}\left [ \frac{1}{\left ( 1-u \right )^3}+\frac{1}{\left ( 1+u \right )^3}+\frac{6}{\left ( 1-u \right )^2} \right ]du\]

\[=\frac{1}{8}\int_{0}^{\frac{\sqrt{2}}{2}}\left [ \frac{1}{\left ( 1-u \right )^3}+\frac{1}{\left ( 1+u \right )^3} \right ]du+\frac{3K}{4}\]

\[=\left.\begin{matrix} \frac{1}{8}\left ( \frac{1}{2\left ( 1-u \right )^2}-\frac{1}{2\left ( 1+u \right )^2} \right ) \end{matrix}\right|_{0}^{\frac{\sqrt{2}}{2}}+\frac{3\sqrt{2}+3ln\left ( 1+\sqrt{2} \right )}{8}=\frac{7\sqrt{2}+3ln\left ( 1+\sqrt{2} \right )}{8}\]

Vậy \[I=H-K=\frac{3\sqrt{2}-ln\left ( 1+\sqrt{2} \right )}{8}\]

Cách 2: 

Đặt \[\left\{\begin{matrix} u=\sqrt{x^2+1}\\dv=x^2dx \end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{xdx}{\sqrt{x^2+1}}\\v=\frac{x^3}{3} \end{matrix}\right.\]

\[\Rightarrow I= \left.\begin{matrix} \frac{x^3}{3}\sqrt{x^2+1} \end{matrix}\right|_{0}^{1}-\int_{0}^{1} \frac{x^4dx}{3\sqrt{x^2+1}}=\frac{\sqrt{2}}{3}-\frac{1}{3}\int_{0}^{1}x^2\sqrt{x^2+1}+\frac{1}{3}\int_{0}^{1}\frac{x^2dx}{\sqrt{x^2}+1}\]

Khi đó ta có \[\frac{4I}{3}=\sqrt{2}+\frac{1}{3}\int_{0}^{1}\frac{x^2dx}{\sqrt{x^2}+1}=\frac{\sqrt{2}}{3}+\frac{J}{3}\]

Trong đó \[J=\int_{0}^{1}\frac{x^2dx}{\sqrt{x^2+1}}\]

Đổi biến \[x=tant,t\in\left ( -\frac{\pi}{2};\frac{\pi}{2} \right )\Rightarrow dx=\left ( tan^2t+1 \right )dt\]

Đổi cận: \[x=0\Rightarrow t=0,x=1\Rightarrow t=\frac{\pi}{4}\]

Khi đó \[J=\int_{0}^{\frac{\pi}{4}}\frac{tan^2t\left ( tan^2t+1 \right )dt}{\sqrt{tan^2t+1}}=\int_{0}^{\frac{\pi}{4}}tan^2t\sqrt{tan^2t+1}dt\]

\[=\int_{0}^{\frac{\pi}{4}}\frac{sin^2tdt}{cos^3t}=\int_{0}^{\frac{\pi}{4}}\frac{sin^2tcostdt}{cos^4t}\]

Đổi biến: \[u=sint, t\in\left [ -1;1 \right ]\Rightarrow du=costdt\]

Đổi cận: \[t=0\Rightarrow u=0, t=\frac{\pi}{4}\Rightarrow u=\frac{\sqrt{2}}{2}\]

Khi đó ta có: \[J=\int_{0}^{\frac{\sqrt{2}}{2}}\frac{u^2du}{\left ( 1 -u^2\right )^2}=\frac{1}{4}\int_{0}^{\frac{\sqrt{2}}{2}}\left ( \frac{1}{u-1}+\frac{1}{u+1} \right )^2du\]

\[=\frac{1}{4}\int_{0}^{\frac{\sqrt{2}}{2}}\left ( \frac{1}{\left (u-1 \right )^2}+\frac{1}{\left (u+1 \right )^2} +\frac{1}{u-1}+\frac{1}{u+1}\right )du\]

\[=\frac{1}{4}\left.\begin{matrix} \left ( -\frac{1}{u-1}-\frac{1}{u+1}+ln\left | u^2-1 \right | \right ) \end{matrix}\right|_{0}^{\frac{\sqrt{2}}{2}}=\frac{\sqrt{2}+ln\left ( \sqrt{2}-1 \right )}{2}\]

Vậy \[I=\frac{3\sqrt{2}+ln\left ( \sqrt{2}-1 \right )}{8}\]

Cách 3: 

\[\left\{\begin{matrix} u=x\\dv=x\sqrt{x^2+1}dx \end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\v=\frac{1}{3}\sqrt{(x^2+1)^3} \end{matrix}\right.\]

\[\Rightarrow I= \left.\begin{matrix} \frac{x\sqrt{\left ( x^2+1 \right )^3}}{3} \end{matrix}\right|_{0}^{1}-\int_{0}^{1} \frac{\left ( x^2+1 \right )\sqrt{x^2+1}dx}{3}\]

\[=\frac{2\sqrt{2}}{3}-\frac{1}{3}\int_{0}^{1}x^2\sqrt{x^2+1}dx-\frac{1}{3}\int_{0}^{1}\sqrt{x^2+1}dx=\frac{2\sqrt{2}}{3}-\frac{I}{3}-\frac{J}{3}\]

Trong đó \[J=\int_{0}^{1}\sqrt{x^2+1}dx\]

Tính J:
Đổi biến: \[x=tant,t\in\left ( -\frac{\pi}{2};\frac{\pi}{2} \right )\Rightarrow dx=(tan^2t+1)dt\]

Đổi cận: \[x=0\Rightarrow t=0,x=1\Rightarrow t=\frac{\pi}{4}\]

Ki đó: \[J=\int_{0}^{\frac{\pi}{4}}(tan^2t+1).\sqrt{tan^2t+1}dt=\int_{0}^{\frac{\pi}{4}}\frac{dt}{cos^3t}\]

Đổi biến: \[u=sint, u\in\left [ -1;1 \right ] \Rightarrow du=costdt\]
Đổi cận: \[t=0\Rightarrow u=0,t=\frac{\pi}{4}\Rightarrow u=\frac{\sqrt{2}}{2}\]
\[J=\int_{0}^{\frac{\sqrt{2}}{2}}\frac{du}{\left ( 1-u^2 \right )^2}\]

Đến đây tính giống K

Vậy \[I=\frac{2\sqrt{2}}{3}-\frac{I}{3}-\frac{J}{3}\Rightarrow I=\frac{3\sqrt{2}+ln(\sqrt{2}-1)}{8}\]

Cách 4: 

\[t=\sqrt{x^2+1},t\geq 1\Rightarrow x^2=t^2-1\Rightarrow x=\sqrt{t^2-1}\]

\[\Rightarrow xdx=tdt\]

Đổi cận: \[x=1\Rightarrow t=1, x=1\Rightarrow t=\sqrt{2}\]

Khi đó \[I=\int_{0}^{\sqrt{2}}t^2\sqrt{t^2-1}dt\]

Đổi biến: \[t=\frac{1}{cosu}\Rightarrow dt=\frac{sinudu}{cos^2u}\]

Đổi cận: \[t=1\Rightarrow u=0, t=\sqrt{2}\Rightarrow u=\frac{\pi}{4}\]

Khi đó \[I=\int_{0}^{\frac{\pi}{4}}\frac{1}{cos^2u}.\sqrt{\frac{1}{cos^2u}-1}.\frac{sinu}{cos^2u}du=\int_{0}^{\frac{\pi}{4}}\frac{sin^2udu}{cos^5u}\]

Đến đây thì hoàn toàn giống với cách 1 ở bước thứ hai.