Primitive of x squared by Root of x squared plus a squared

From ProofWiki
Jump to navigation Jump to search

Theorem

$\displaystyle \int x^2 \sqrt {x^2 + a^2} \rd x = \frac {x \paren {\sqrt {x^2 + a^2} }^3} 4 - \frac {a^2 x \sqrt {x^2 + a^2} } 8 - \frac {a^4} 8 \map \ln {x + \sqrt {x^2 + a^2} } + C$


Proof

Let:

\(\displaystyle z\) \(=\) \(\displaystyle x^2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\d z} {\d x}\) \(=\) \(\displaystyle 2 x\) Power Rule for Derivatives
\(\displaystyle \leadsto \ \ \) \(\displaystyle \int x^2 \sqrt {x^2 + a^2} \rd x\) \(=\) \(\displaystyle \int \frac {z \sqrt {z + a^2} \rd z} {2 \sqrt z}\) Integration by Substitution
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 \int \sqrt z \sqrt {z + a^2} \rd z\) Primitive of Constant Multiple of Function
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 \paren {\frac {2 z + a^2} 4 \sqrt z \sqrt {z + a^2} - \frac {a^4} 8 \int \frac {\d z} {\sqrt z \sqrt {z + a^2} } } + C\) Primitive of $\sqrt {\paren {a x + b} \paren {p x + q} }$
\(\displaystyle \) \(=\) \(\displaystyle \frac {2 z + a^2} 8 \sqrt z \sqrt {z + a^2} - \frac {a^4} {16} \paren {2 \map \ln {\sqrt {z + a^2} + \sqrt z} } + C\) Primitive of $\dfrac 1 {\sqrt {\paren {a x + b} \paren {p x + q} } }$
\(\displaystyle \) \(=\) \(\displaystyle \frac {2 z + 2 a^2} 8 \sqrt z \sqrt {z + a^2} - \frac {a^2} 8 \sqrt z \sqrt {z + a^2} - \frac {a^4} 8 \map \ln {\sqrt {z + a^2} + \sqrt z} + C\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {z + a^2} 4 \sqrt z \sqrt {z + a^2} - \frac {a^2} 8 \sqrt z \sqrt {z + a^2} - \frac {a^4} 8 \map \ln {\sqrt {z + a^2} + \sqrt z} + C\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {x \paren {\sqrt {x^2 + a^2} }^3} 4 - \frac {a^2 x \sqrt {x^2 + a^2} } 8 - \frac {a^4} 8 \map \ln {x + \sqrt {x^2 + a^2} } + C\) substituting for $z$

$\blacksquare$


Also see


Sources