# Pi as Sum of Sequence of Reciprocal of Product of Three Consecutive Integers/Lemma

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## Theorem

- $\displaystyle \iiint \dfrac x {x^2 + 1} \rd x \rd x \rd x = x \map \arctan x + \dfrac {\paren {x^2 - 1} \map \ln {x^2 + 1} - 3 x^2} 4$

with all integration constants at $0$.

## Proof

First primitive:

\(\displaystyle \int \dfrac x {x^2 + 1} \rd x\) | \(=\) | \(\displaystyle \dfrac 1 2 \int \dfrac {2 x} {x^2 + 1} \rd x\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac {\map \ln {x^2 + 1} } 2\) | Primitive of Function under its Derivative |

The integration constant is not added due to the series never having a constant during its integration.

$\Box$

Second primitive:

\(\displaystyle \int \dfrac {\map \ln {x^2 + 1} } 2 \rd x\) | \(=\) | \(\displaystyle \dfrac 1 2 \paren {x \map \ln {x^2 + 1^2} - 2 x + 2 \times 1 \times \arctan \frac x 1}\) | Primitive of $\ln {x^2 + a^2}$, putting $a = 1$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac {x \map \ln {x^2 + 1} } 2 - x + \map \arctan x\) |

$\Box$

Third primitive:

\(\displaystyle \int \paren {\dfrac {x \map \ln {x^2 + 1} } 2 + \map \arctan x - x} \rd x\) | \(=\) | \(\displaystyle \int \dfrac {x \map \ln {x^2 + 1} } 2 \rd x + \int \map \arctan x \rd x - \int x \rd x\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \int \dfrac {x \map \ln {x^2 + 1} } 2 \rd x + \int \map \arctan x \rd x - \dfrac {x^2} 2\) | Integral of Power | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \int \dfrac {x \map \ln {x^2 + 1} } 2 \rd x + x \map \arctan x - \dfrac {\map \ln {x^2 + 1} } 2 - \dfrac {x^2} 2\) | Primitive of $\arctan \dfrac x a$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 2 \paren {\frac {\paren {x^2 + 1} \map \ln {x^2 + 1} - x^2} 2} + x \map \arctan x - \dfrac {\map \ln {x^2 + 1} } 2 - \dfrac {x^2} 2\) | Primitive of $x \map \ln {x^2 + a^2}$ with $a = 1$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x \map \arctan x + \dfrac {\paren {x^2 - 1} \map \ln {x^2 + 1} - 3 x^2} 4\) | simplifying |

$\blacksquare$