Primitive of x cubed over a squared minus x squared squared

From ProofWiki
Jump to navigation Jump to search

Theorem

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

for $x^2 < a^2$.


Proof

\(\ds \int \frac {x^3 \rd x} {\paren {a^2 - x^2}^2}\) \(=\) \(\ds \int \frac {x \paren {x^2 - a^2 + a^2} } {\paren {a^2 - x^2}^2} \rd x\)
\(\ds \) \(=\) \(\ds \int \frac {-x \paren {a^2 - x^2} } {\paren {a^2 - x^2}^2} \rd x + a^2 \int \frac {x \rd x} {\paren {a^2 - x^2}^2}\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \int \frac {-x \rd x} {a^2 - x^2} + a^2 \int \frac {x \rd x} {\paren {a^2 - x^2}^2}\) simplification
\(\ds \) \(=\) \(\ds -\frac 1 2 \paren {-\map \ln {a^2 - x^2} } + a^2 \int \frac {x \rd x} {\paren {a^2 - x^2}^2} + C\) Primitive of $\dfrac x {a^2 - x^2}$
\(\ds \) \(=\) \(\ds \frac 1 2 \map \ln {a^2 - x^2} + a^2 \paren {\frac 1 {2 \paren {a^2 - x^2} } } + C\) Primitive of $\dfrac x {\paren {a^2 - x^2}^2}$
\(\ds \) \(=\) \(\ds \frac {a^2} {2 \paren {a^2 - x^2} } + \frac 1 2 \map \ln {a^2 - x^2} + C\) simplifying

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Primitive_of_x_cubed_over_a_squared_minus_x_squared_squared&oldid=598138"