Primitive of x by Logarithm of x squared plus a squared
Jump to navigation
Jump to search
Theorem
- $\ds \int x \map \ln {x^2 + a^2} \rd x = \frac {\paren {x^2 + a^2} \map \ln {x^2 + a^2} - x^2} 2 + C$
Proof 1
| \(\ds \int x \map \ln {x^2 + a^2} \rd x\) | \(=\) | \(\ds \frac {x^2 \map \ln {x^2 + a^2} } 2 - \int \frac {x^3} {x^2 + a^2} \rd x + C\) | Primitive of $x^m \map \ln {x^2 + a^2}$ with $m = 1$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {x^2 \map \ln {x^2 + a^2} } 2 - \paren {\frac {x^2} 2 - \frac {a^2} 2 \map \ln {x^2 + a^2} } + C\) | Primitive of $\dfrac {x^3} {x^2 + a^2}$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\paren {x^2 + a^2} \map \ln {x^2 + a^2} - x^2} 2 + C\) | simplifying |
$\blacksquare$
Proof 2
Let $u := x^2 + a^2$.
Then:
| \(\ds \dfrac {\d u} {\d x}\) | \(=\) | \(\ds 2 x\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \int x \map \ln {x^2 + a^2} \rd x\) | \(=\) | \(\ds \int \dfrac 1 2 \dfrac {\d u} {\d x} \ln u \rd x\) | Integration by Substitution | ||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 2 \int \ln u \rd u\) | simplifying | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 2 \paren {u \ln u - u + C}\) | Primitive of $\ln u$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 2 \paren {\paren {x^2 + a^2} \map \ln {x^2 + a^2} - x^2 + a^2} + C\) | substituting back | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {\paren {x^2 + a^2} \map \ln {x^2 + a^2} - x^2} 2 + C\) | subsuming $\dfrac {a^2} 2$ into the arbitrary constant and simplifying |
$\blacksquare$