Indefinite integral (02)

Calculus (Integrals, Series)
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Grigorios Kostakos
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Indefinite integral (02)

#1

Post by Grigorios Kostakos »

Find \[\displaystyle\int{\frac{1+x^2}{({1-x^2})\,\sqrt{1+x^4}}\,dx}\,.\]
Grigorios Kostakos
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Re: Indefinite integral (02)

#2

Post by Grigorios Kostakos »

1st solution:
\begin{align*}
& \displaystyle\int{\frac{1+x^2}{({1-x^2})\sqrt{1+x^4}}\,dx}=\int{\frac{({1+x^2})\,({1-x^2})}{\left({1-x^2}\right)^2\sqrt{1+x^4}}\,dx}=\\
& \displaystyle\int{\frac{1-x^4}{\bigl({\sqrt{1+x^4}+x\,\sqrt{2}\,}\bigr)\bigl({\sqrt{1+x^4}-x\,\sqrt{2}\,}\bigr)\sqrt{1+x^4}}\,dx}=\\
&\displaystyle\frac{1}{2\sqrt{2}}\int{\frac{\bigl({\sqrt{1+x^4}-x\,\sqrt{2}\,}\bigr)\bigl({2x^3+\sqrt{2}\,\sqrt{1+x^4}\,}\bigr)-\bigl({\sqrt{1+x^4}-x\,\sqrt{2}\,}\bigr)\bigl({2x^3-\sqrt{2}\,\sqrt{1+x^4}\,}\bigr)}{\bigl({\sqrt{1+x^4}+x\,\sqrt{2}\,}\bigr)\bigl({\sqrt{1+x^4}-x\,\sqrt{2}\,}\bigr)\sqrt{1+x^4}}\,dx}=\\
& \displaystyle\frac{1}{2\sqrt{2}}\int{\frac{2x^3+\sqrt{2}\,\sqrt{1+x^4}}{\bigl({\sqrt{1+x^4}+x\,\sqrt{2}\,}\bigr)\,\sqrt{1+x^4}}\,dx}-\frac{1}{2\sqrt{2}}\int{\frac{2x^3-\sqrt{2}\,\sqrt{1+x^4}}{\bigl({\sqrt{1+x^4}-x\,\sqrt{2}\,}\bigr)\,\sqrt{1+x^4}}\,dx}=\\
& \displaystyle\frac{1}{2\sqrt{2}}\int{\frac{\frac{2x^3}{\sqrt{1+x^4}}+\sqrt{2}}{\sqrt{1+x^4}+x\,\sqrt{2}}\,dx}-\frac{1}{2\sqrt{2}}\int{\frac{\frac{2x^3}{\sqrt{1+x^4}}-\sqrt{2}}{\sqrt{1+x^4}-x\,\sqrt{2}}\,dx}=\\
& \displaystyle\frac{1}{2\sqrt{2}}\int{\frac{\bigl({\sqrt{1+x^4}+x\,\sqrt{2}\,}\bigr)^{\prime}}{\sqrt{1+x^4}+x\,\sqrt{2}}\,dx}-\frac{1}{2\sqrt{2}}\int{\frac{\bigl({\sqrt{1+x^4}-x\,\sqrt{2}\,}\bigr)^{\prime}}{\sqrt{1+x^4}-x\,\sqrt{2}}\,dx}=\\
& \displaystyle\frac{1}{2\sqrt{2}}\,\log\bigl|{\sqrt{1+x^4}+x\,\sqrt{2}\,}\bigr|-\frac{1}{2\sqrt{2}}\,\log\bigl|{\sqrt{1+x^4}-x\,\sqrt{2}\,}\bigr|+c=\frac{1}{2\sqrt{2}}\,\log\left|{\tfrac{\sqrt{1+x^4}+x\,\sqrt{2}}{\sqrt{1+x^4}-x\,\sqrt{2}}}\right|+c\,.
\end{align*}
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Re: Indefinite integral (02)

#3

Post by Grigorios Kostakos »

2nd solution:
\(\displaystyle\int{\frac{1+x^2}{\bigl({1-x^2}\bigr)\sqrt{1+x^4}}\,dx}=\int{\frac{\bigl({\sqrt{1+x^4}+x\,\sqrt{2}\,}\bigr)\,({1+x^2})}{\bigl({\sqrt{1+x^4}+x\,\sqrt{2}\,}\bigr)\,({1-x^2})\,\sqrt{1+x^4}}\,dx}=\)

\(\displaystyle\int{\frac{\sqrt{1+x^4}+x^2\,\sqrt{1+x^4}+x\,\sqrt{2}+x^3\,\sqrt{2}}{\bigl({\sqrt{1+x^4}+x\,\sqrt{2}\,}\bigr)\,({1-x^2})\,\sqrt{1+x^4}}\,dx}=\)

\(\displaystyle\frac{1}{\sqrt{2}}\int{\frac{\sqrt{2}\,\sqrt{1+x^4}+x^2\,\sqrt{2}\,\sqrt{1+x^4}+2x+2x^3}{\bigl({\sqrt{1+x^4}+x\,\sqrt{2}\,}\bigr)\,({1-x^2})\,\sqrt{1+x^4}}\,dx}=\)

\(\displaystyle\frac{1}{\sqrt{2}}\int{\frac{2x^5+2x^2\,\sqrt{2}\,\sqrt{1+x^4}+2x-2x^5+2x^3-x^2\,\sqrt{2}\,\sqrt{1+x^4}+\sqrt{2}\,\sqrt{1+x^4}}{\bigl({\sqrt{1+x^4}+x\,\sqrt{2}\,}\bigr)\,({1-x^2})\,\sqrt{1+x^4}}\,dx}=\)

\(\displaystyle\frac{1}{\sqrt{2}}\int{\frac{2x\,\bigl({x^4+2x\,\sqrt{2}\,\sqrt{1+x^4}+1}\bigr)}{\bigl({\sqrt{1+x^4}+x\,\sqrt{2}\,}\bigr)\,({1-x^2})\,\sqrt{1+x^4}}\,dx}+\displaystyle\frac{1}{\sqrt{2}}\int{\frac{({1-x^2})\,\bigl({\sqrt{2}\,\sqrt{1+x^4}+2x^3}\bigr)}{\bigl({\sqrt{1+x^4}+x\,\sqrt{2}\,}\bigr)\,({1-x^2})\,\sqrt{1+x^4}}\,dx}=\)

\(\displaystyle\frac{1}{\sqrt{2}}\int{\frac{2x\,\sqrt{1+x^4}\,\bigl({\sqrt{1+x^4}+x\,\sqrt{2}\,}\bigr)}{\bigl({\sqrt{1+x^4}+x\,\sqrt{2}\,}\bigr)\,({1-x^2})\,\sqrt{1+x^4}}\,dx}+\frac{1}{\sqrt{2}}\int{\frac{\sqrt{2}\,\sqrt{1+x^4}+2x^3}{\bigl({\sqrt{1+x^4}+x\,\sqrt{2}\,}\bigr)\,({1-x^2})\,\sqrt{1+x^4}}\,dx}=\)

\(\displaystyle\frac{1}{\sqrt{2}}\int{\frac{2x}{1-x^2}\,dx}+\frac{1}{\sqrt{2}}\int{\frac{\frac{\sqrt{2}\,\sqrt{1+x^4}+2x^3}{\sqrt{1+x^4}}}{\sqrt{1+x^4}+x\,\sqrt{2}}\,dx}=\)

\(\displaystyle-\frac{1}{\sqrt{2}}\int{\frac{({1-x^2})^{\prime}}{1-x^2}\,dx}+\frac{1}{\sqrt{2}}\int{\frac{\sqrt{2}+\frac{2x^3}{\sqrt{1+x^4}}}{\sqrt{1+x^4}+x\,\sqrt{2}}\,dx}=\)

\(\displaystyle-\frac{1}{\sqrt{2}}\,\log\bigl|{1-x^2}\bigr|+c_1+\frac{1}{\sqrt{2}}\int{\frac{\bigl({\sqrt{1+x^4}+x\,\sqrt{2}\,}\bigr)^{\prime}}{\sqrt{1+x^4}+x\,\sqrt{2}}\,dx}=\)

\(\displaystyle-\frac{1}{\sqrt{2}}\,\log\left|{1-x^2}\right|+\frac{1}{\sqrt{2}}\,\log\bigl|{\sqrt{1+x^4}+x\,\sqrt{2}\,}\bigr|+c=\frac{1}{\sqrt{2}}\,\log\left|{\tfrac{\sqrt{1+x^4}+x\,\sqrt{2}}{1-x^2}}\right|+c=\)

\(\displaystyle\frac{1}{2\sqrt{2}}\,\log\left|{\tfrac{\left({\sqrt{1+x^4}+x\,\sqrt{2}}\right)^2}{\left({\sqrt{1+x^4}+x\,\sqrt{2}}\right)\left({\sqrt{1+x^4}-x\,\sqrt{2}}\right)}}\right|+c=\frac{1}{2\sqrt{2}}\,\log\left|{\tfrac{\sqrt{1+x^4}+x\,\sqrt{2}}{\sqrt{1+x^4}-x\,\sqrt{2}}}\right|+c\,.\quad\square\)
Grigorios Kostakos
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Re: Indefinite integral (02)

#4

Post by jacks »

Given \(\displaystyle\int\frac{1+x^2}{\left(1-x^2\right)\sqrt{1+x^4}}dx\)

Divide both \(N_{r}\) and \(D_{r}\) by \(x^2\)

\begin{align*}
&=-\int\frac{\left(1+\frac{1}{x^2}\right)}{\left(x-\frac{1}{x}\right)\sqrt{x^2+\frac{1}{x^2}}}dx\\
&=-\int\frac{\left(1+\frac{1}{x^2}\right)}{\left(x-\frac{1}{x}\right)\sqrt{\left(x-\frac{1}{x}\right)^2+2}}dx
\end{align*}
Now Put \(\left(x-\frac{1}{x}\right)=t\quad\Leftrightarrow\quad\left(1+\frac{1}{x^2}\right)dx\,=\,dt\)

\[=-\int\frac{1}{t\sqrt{t^2+2}}dt\]

Now Let \(t^2+2=u^2\quad\Leftrightarrow\quad tdt=udu\quad\Leftrightarrow\quad dt=\frac{u}{t}dt\)

\begin{align*}
&=-\int\frac{1}{u^2-\left(\sqrt{2}\right)^2}du=\int\frac{1}{\left(\sqrt{2}\right)^2-u^2}du\\
&=\frac{1}{2\sqrt{2}}\ln\left|\frac{\sqrt{2}+u}{\sqrt{2}-u}\right|+C\\
&=\frac{1}{2\sqrt{2}}\ln\left|\frac{\sqrt{2}+\sqrt{t^2+2}}{\sqrt{2}-\sqrt{t^2+2}}\right|+C
\end{align*}
So \[\int\frac{1+x^2}{\left(1-x^2\right)\sqrt{1+x^4}}dx=\frac{1}{2\sqrt{2}}\ln\left|\frac{\sqrt{2}x+\sqrt{1+x^4}}{\sqrt{2}x-\sqrt{1+x^4}}\right|+C\,.\]
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