Primitive of Power of x by Arctangent of x over a/Proof 1
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Theorem
- $\ds \int x^m \arctan \frac x a \rd x = \frac {x^{m + 1} } {m + 1} \arctan \frac x a - \frac a {m + 1} \int \frac {x^{m + 1} \rd x} {x^2 + a^2}$
Proof
Recall:
| \(\text {(1)}: \quad\) | \(\ds \int x^m \arctan x \rd x\) | \(=\) | \(\ds \frac {x^{m + 1} } {m + 1} \arctan x - \frac 1 {m + 1} \int \frac {x^{m + 1} \rd x} {x^2 + 1}\) | Primitive of $x^m \arctan x$ |
Then:
| \(\ds \int x^m \arctan \frac x a \rd x\) | \(=\) | \(\ds \int a^m \paren {\dfrac x a}^m \arctan \frac x a \rd x\) | manipulating into appropriate form | |||||||||||
| \(\ds \) | \(=\) | \(\ds a^m \int \paren {\dfrac x a}^m \arctan \frac x a \rd x\) | Primitive of Constant Multiple of Function | |||||||||||
| \(\ds \) | \(=\) | \(\ds a^m \paren {\dfrac 1 {1 / a} \paren {\frac 1 {m + 1} \paren {\dfrac x a}^{m + 1} \arctan \frac x a - \frac 1 {m + 1} \int \paren {\dfrac x a}^{m + 1} \frac {\d x} {\paren {\dfrac x a}^2 + 1} } }\) | Primitive of Function of Constant Multiple, from $(1)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {x^{m + 1} } {m + 1} \arctan \frac x a - \frac a {m + 1} \int \frac {x^{m + 1} \rd x} {x^2 + a^2}\) | simplifying |
$\blacksquare$