Primitive of Arctangent of x over a over x squared

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Theorem

$\displaystyle \int \frac {\arctan \frac x a \ \mathrm d x} {x^2} = \frac {-\arctan \frac x a} x - \frac 1 {2 a} \ln \left({\frac {x^2 + a^2} {x^2} }\right) + C$


Proof

With a view to expressing the primitive in the form:

$\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \ \mathrm d x = u v - \int v \frac {\mathrm d u}{\mathrm d x} \ \mathrm d x$

let:

\(\displaystyle u\) \(=\) \(\displaystyle \arctan \frac x a\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\mathrm d u} {\mathrm d x}\) \(=\) \(\displaystyle \frac a {x^2 + a^2}\) Derivative of $\arctan \dfrac x a$


and let:

\(\displaystyle \frac {\mathrm d v} {\mathrm d x}\) \(=\) \(\displaystyle \frac 1 {x^2}\)
\(\displaystyle \implies \ \ \) \(\displaystyle v\) \(=\) \(\displaystyle \frac {-1} x\) Primitive of Power


Then:

\(\displaystyle \int \frac {\arctan \frac x a \ \mathrm d x} {x^2}\) \(=\) \(\displaystyle \arctan \frac x a \left({\frac {-1} x}\right) - \int \left({\frac {-1} x}\right) \left({\frac a {x^2 + a^2} }\right) \ \mathrm d x + C\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\arctan \frac x a} x + a \int \frac {\mathrm d x} {x \left({x^2 + a^2}\right)} \ \mathrm d x + C\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\arctan \frac x a} x + \frac 1 a \left({\frac 1 {2 a^2} \ln \left({\frac {x^2} {x^2 + a^2} }\right)}\right) + C\) Primitive of $\dfrac 1 {x \left({x^2 + a^2}\right)}$
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\arctan \frac x a} x - \frac 1 {2 a} \ln \left({\frac {x^2 + a^2} {x^2} }\right) + C\) Logarithm of Reciprocal

$\blacksquare$


Also see


Sources