Primitive of x cubed over x squared minus a squared

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Theorem

$\displaystyle \int \frac {x^3 \ \mathrm d x} {x^2 - a^2} = \frac {x^2} 2 + \frac {a^2} 2 \ln \left({x^2 - a^2}\right) + C$

for $x^2 > a^2$.


Proof

Let:

\(\displaystyle \int \frac {x^3 \ \mathrm d x} {x^2 - a^2}\) \(=\) \(\displaystyle \int \frac {x \left({x^2 - a^2 + a^2}\right)} {x^2 - a^2} \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle \int \frac {x \left({x^2 - a^2}\right)} {x^2 - a^2} \ \mathrm d x + \int \frac {a^2 x} {x^2 - a^2} \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle \int x \ \mathrm d x + a^2 \int \frac {x \ \mathrm d x} {x^2 - a^2}\) Primitive of Constant Multiple of Function
\(\displaystyle \) \(=\) \(\displaystyle \frac {x^2} 2 + a^2 \int \frac {x \ \mathrm d x} {x^2 - a^2} + C\) Primitive of Power
\(\displaystyle \) \(=\) \(\displaystyle \frac {x^2} 2 + a^2 \left({\frac 1 2 \ln \left({x^2 - a^2}\right)}\right) + C\) Primitive of $\dfrac x {x^2 - a^2}$
\(\displaystyle \) \(=\) \(\displaystyle \frac {x^2} 2 + \frac {a^2} 2 \ln \left({x^2 - a^2}\right) + C\) simplifying

$\blacksquare$


Also see


Sources