Primitive of x squared over x cubed plus a cubed

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Theorem

$\ds \int \frac {x^2 \rd x} {x^3 + a^3} = \frac 1 3 \ln \size {x^3 + a^3} + C$


Proof 1

\(\ds \map {\frac \d {\d x} } {x^3 + a^3}\) \(=\) \(\ds 3 x^2\) Derivative of Power
\(\ds \leadsto \ \ \) \(\ds \int \frac {x^2 \rd x} {x^3 + a^3}\) \(=\) \(\ds \frac 1 3 \ln \size {x^3 + a^3} + C\) Primitive of Function under its Derivative

$\blacksquare$


Proof 2

From Primitive of Power of x less one over Power of x plus Power of a:

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


So:

\(\ds \int \frac {x^2 \rd x} {x^3 + a^3}\) \(=\) \(\ds \frac 1 3 \ln \size {x^3 + a^3} + C\) Primitive of $\dfrac {x^{n - 1} } {\paren {x^n + a^n} }$ with $n = 3$

directly.

$\blacksquare$


Sources

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