Yankel said:
Hello all
I am working on this integral
\[\int \frac{x^{2}+1}{x^{4}+1}dx\]Now, I have tried this way:
\[u=x^{2}+1\]
after I did:
\[\int \frac{x^{2}+1}{\left ( x^{2}+1 \right )\left ( x^{2}-1 \right )}dx\]
But I got stuck, I got:
\[\frac{1}{2}\cdot \int \frac{1}{u\sqrt{u-1}}dx\]
I thought of making another substitution, but I tried and failed. Help required
Well first of all, [math]\displaystyle \begin{align*} x^4 + 1 \end{align*}[/math] is NOT equal to [math]\displaystyle \begin{align*} \left( x^2 + 1 \right) \left( x^2 - 1 \right) \end{align*}[/math].
I would try
[math]\displaystyle \begin{align*} x^4 + 1 &= x^4 + 2x^2 + 1 - 2x^2 \\ &= \left( x^2 + 1 \right) ^2 - \left( \sqrt{2} \, x \right) ^2 \\ &= \left( x^2 - \sqrt{2}\, x + 1 \right) \left( x^2 + \sqrt{2}\,x + 1 \right) \end{align*}[/math]
and so
[math]\displaystyle \begin{align*} \int{ \frac{x^2 + 1}{x^4 + 1}\,dx} &= \int{ \frac{x^2 - \sqrt{2}\,x + 1 + \sqrt{2}\,x}{ \left( x^2 - \sqrt{2}\,x + 1 \right) \left( x^2 + \sqrt{2}\,x + 1 \right) } \, dx} \\ &= \int{ \frac{1}{x^2 + \sqrt{2}\,x + 1} \,dx} + \sqrt{2} \int{ \frac{x}{ \left( x^2 - \sqrt{2}\,x + 1 \right) \left( x^2 + \sqrt{2}\,x + 1 \right) } \, dx} \\ &= \int{ \frac{1}{x^2 + \sqrt{2}\,x + \left( \frac{\sqrt{2}}{2} \right) ^2 - \left( \frac{\sqrt{2}}{2} \right) ^2 + 1 } \, dx} + \sqrt{2} \int{ \frac{x}{x^4 + 1} \, dx} \\ &= \int{ \frac{1}{ \left( x + \frac{\sqrt{2}}{2} \right) ^2 + \frac{1}{2} } \, dx} + \frac{\sqrt{2}}{2} \int{ \frac{2x}{ \left( x^2 \right) ^2 + 1 }\, dx} \end{align*}[/math]
Now in the first integral, let [math]\displaystyle \begin{align*} x + \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}\tan{(\theta)} \implies dx = \frac{\sqrt{2}}{2}\sec^2{(\theta)}\,d\theta \end{align*}[/math] and in the second integral let [math]\displaystyle \begin{align*} x^2 = \tan{(\phi)} \implies 2x\,dx = \sec^2{(\phi)}\,d\phi \end{align*}[/math] and the integrals become
[math]\displaystyle \begin{align*} \int{ \frac{1}{\left( x + \frac{\sqrt{2}}{2} \right) ^2 + \frac{1}{2} } \, dx} + \frac{\sqrt{2}}{2} \int{ \frac{2x}{ \left( x^2 \right) ^2 + 1 } \, dx} &= \int{ \frac{1}{ \left[ \frac{\sqrt{2}}{2} \tan{(\theta)} \right] ^2 + \frac{1}{2} } \cdot \frac{\sqrt{2}}{2}\sec^2{(\theta)}\,d\theta } + \frac{\sqrt{2}}{2} \int{ \frac{\sec^2{(\phi)}}{ \tan^2{(\phi)} + 1 } \, d\phi } \\ &= \frac{\sqrt{2}}{2} \int{ \frac{\sec^2{(\theta)}}{ \frac{1}{2}\tan^2{(\theta)} + \frac{1}{2} } \, d\theta} + \frac{\sqrt{2}}{2} \int{ \frac{\sec^2{(\phi)}}{\sec^2{(\phi)}}\,d\phi} \\ &= \sqrt{2} \int{ \frac{\sec^2{(\theta)}}{ \tan^2{(\theta)} + 1 } \, d\theta} + \frac{\sqrt{2}}{2} \int{ 1 \, d\phi } \\ &= \sqrt{2} \int{ \frac{\sec^2{(\theta)}}{\sec^2{(\theta)}}\,d\theta} + \frac{\sqrt{2}}{2} \phi + C_1 \\ &= \sqrt{2} \int{ 1\,d\theta } + \frac{\sqrt{2}}{2} \arctan{ \left( x^2 \right) } + C_1 \\ &= \sqrt{2} \, \theta + C_2 + \frac{\sqrt{2}}{2}\arctan{ \left( x^2 \right) } + C_1 \\ &= \sqrt{2} \arctan{ \left( \sqrt{2}\, x + 1 \right) } + \frac{\sqrt{2}}{2} \arctan{ \left( x^2 \right) } + C \textrm{ where } C = C_1 + C_2 \end{align*}[/math]
