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

## 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

$\blacksquare$