Primitive of x cubed over x fourth plus a fourth/Proof 2

Theorem

$\ds \int \frac {x^3 \rd x} {x^4 + a^4} = \frac {\map \ln {x^4 + a^4} } 4 + C$

Proof

From Primitive of $\dfrac {x^{n - 1} } {x^n + a^n}$:

$\ds \int \frac {x^{n - 1} \rd x} {x^n + a^n} = \frac 1 n \ln \size {x^n + a^n} + C$

So:

\(\ds \int \frac {x^3 \rd x} {x^4 + a^4}\)

\(=\)

\(\ds \frac 1 4 \ln \size {x^4 + a^4} + C\)

Primitive of $\dfrac {x^{n - 1} } {\paren {x^n + a^n} }$ with $n = 4$

\(\ds \)

\(=\)

\(\ds \frac {\map \ln {x^4 + a^4} } 4 + C\)

Absolute Value of Even Power

$\blacksquare$