Primitive of Minus One plus x Squared over One plus Fourth Power of x
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Theorem
- $\ds \int \frac {x^2 - 1} {x^4 + 1} \rd x = \frac 1 {2 \sqrt 2} \ln \size {\frac {x^2 - \sqrt 2 x + 1} {x^2 + \sqrt 2 x + 1} } + C$
Proof 1
We have:
\(\ds \int \frac {x^2 - 1} {x^4 + 1} \rd x\) | \(=\) | \(\ds \int \frac {1 - \frac 1 {x^2} } {x^2 + \frac 1 {x^2} } \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {1 - \frac 1 {x^2} } {\paren {x + \frac 1 x}^2 - 2} \rd x\) | Completing the Square |
Note that, by Derivative of Power:
- $\dfrac \d {\d x} \paren {x + \dfrac 1 x} = 1 - \dfrac 1 {x^2}$
So, we have:
\(\ds \int \frac {1 - \frac 1 {x^2} } {\paren {x + \frac 1 x}^2 - 2} \rd x\) | \(=\) | \(\ds \int \frac 1 {u^2 - 2} \rd u\) | substituting $u = x + \dfrac 1 x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 \sqrt 2} \ln \size {\frac {u - \sqrt 2} {u + \sqrt 2} } + C\) | Primitive of $\dfrac 1 {x^2 - a^2}$: Logarithm Form | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 \sqrt 2} \ln \size {\frac {x + \frac 1 x - \sqrt 2} {x + \frac 1 x + \sqrt 2} } + C\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 \sqrt 2} \ln \size {\frac {x^2 - \sqrt 2 x + 1} {x^2 + \sqrt 2 x + 1} } + C\) |
$\blacksquare$
Proof 2
\(\ds \int \frac {x^2 - 1} {x^4 + 1} \rd x\) | \(=\) | \(\ds \int \frac {x^2} {x^4 + 1} \rd x - \int \frac 1 {x^4 + 1} \rd x\) | Linear Combination of Primitives | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {4 \sqrt 2} \map \ln {\frac {x^2 - x \sqrt 2 + 1} {x^2 + x \sqrt 2 + 1} } - \frac 1 {2 \sqrt 2} \paren {\map \arctan {1 - x \sqrt 2} - \map \arctan {1 + x \sqrt 2} }\) | Primitive of $\dfrac {x^2} {x^4 + a^4}$, setting $a = 1$ | |||||||||||
\(\ds \) | \(\) | \(\, \ds - \, \) | \(\ds \frac 1 {2 \sqrt 2} \paren {\map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } + \frac 1 2 \ln \size {\frac {x^2 + \sqrt 2 x + 1} {x^2 - \sqrt 2 x + 1} } }\) | Primitive of $\dfrac 1 {1 + x^4}$ |
This needs considerable tedious hard slog to complete it. In particular: Resolve the arctangent equality To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |