Primitive of Reciprocal of x squared plus a squared/Arctangent Form

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Theorem

$\ds \int \frac {\d x} {x^2 + a^2} = \frac 1 a \arctan \frac x a + C$

where $a$ is a non-zero constant.


Proof 1

Let:

$a \tan \theta = x$

for $\theta \in \openint {-\dfrac \pi 2} {\dfrac \pi 2}$.

From Shape of Tangent Function, this substitution is valid for all real $x$.

Then:

\(\ds x\) \(=\) \(\ds a \tan \theta\) from above
\(\ds \leadsto \ \ \) \(\ds \frac {\d x} {\d \theta}\) \(=\) \(\ds a \sec^2 \theta\) Derivative of Tangent Function
\(\ds \leadsto \ \ \) \(\ds \int \frac 1 {x^2 + a^2} \rd x\) \(=\) \(\ds \int \frac {a \ \sec^2 \theta} {a^2 \tan^2 \theta + a^2} \rd \theta\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac a {a^2} \int \frac {\sec^2 \theta} {\tan^2 \theta + 1} rd \theta\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds \frac 1 a \int \frac {\sec^2 \theta} {\sec^2 \theta} \rd \theta\) Difference of Squares of Secant and Tangent
\(\ds \) \(=\) \(\ds \frac 1 a \int \rd \theta\)
\(\ds \) \(=\) \(\ds \frac 1 a \theta + C\) Integral of Constant


As $\theta$ was stipulated to be in the open interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$:

$\tan \theta = \dfrac x a \iff \theta = \arctan \dfrac x a$

Thus:

$\displaystyle \int \frac 1 {x^2 + a^2} \rd x = \frac 1 a \arctan \frac x a + C$

$\Box$


When $a = 0$, both $\dfrac x a$ and $\dfrac 1 a$ are undefined.

However, consider the limit of the above primitive as $a \to 0$:

\(\ds \lim_{a \mathop \to 0} \frac 1 a \arctan {\frac x a}\) \(=\) \(\ds \lim_{a \mathop \to 0} \frac {\arctan {\frac x a} } a\)
\(\ds \leadsto \ \ \) \(\ds \) \(=\) \(\ds \lim_{a \mathop \to 0} \frac {-x a^{-2} } {1 + \frac {x^2} {a^2} }\) L'Hôpital's Rule, Derivative of Arctangent Function
\(\ds \) \(=\) \(\ds \lim_{a \mathop \to 0} \frac {a^{-2} } {a^{-2} } \frac {-x} {x^2 + a^2}\)
\(\ds \) \(=\) \(\ds -\frac 1 x\)

This corresponds with the result:

$\displaystyle \int \frac 1 {x^2} \rd x = \frac {-1} x + C$

which follows from Primitive of Power.

$\blacksquare$


Proof 2

We have that $x^2 + a^2$ is in the form $a x^2 + b x + c$, where $b^2 - 4 a c < 0$.

Thus from Primitive of $\dfrac 1 {a x^2 + b x + c}$ for $b^2 - 4 a c > 0$:

$\ds \int \frac {\d x} {a x^2 + b x + c} = \frac 2 {\sqrt {4 a c - b^2} } \map \arctan {\frac {2 a x + b} {\sqrt {4 a c - b^2} } } + C$

setting $a := 1, b := 0, c := a^2$:

\(\ds \int \frac 1 {x^2 + a^2} \rd x\) \(=\) \(\ds \dfrac 2 {\sqrt {4 a^2 - 0} } \map \arctan {\dfrac {2 x + 0} {\sqrt {4 a^2} } } + C\) Primitive of $\dfrac 1 {a x^2 + b x + c}$
\(\ds \) \(=\) \(\ds \frac 1 a \arctan {\frac x a} + C\) simplifying

$\blacksquare$


Proof 3

\(\ds \int \frac {\d x} {x^2 + a^2}\) \(=\) \(\ds \frac 1 a \int \frac {\d t} {t^2 + 1}\) Substitution of $x \to a t$
\(\ds \) \(=\) \(\ds \frac 1 a \int \frac {\d t} {\paren {1 + i t} \paren {1 - i t} }\) Factoring
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \paren {\int \frac {\d t} {1 + i t} + \int \frac {\d t} {1 - i t} }\) Definition of Partial Fractions Expansion
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \paren {i \map \ln {1 - i t} - i \map \ln {1 + i t} } + C\) Primitive of Reciprocal
\(\ds \) \(=\) \(\ds \frac i {2 a} \map \ln {\frac {1 - i t} {1 + i t} } + C\) Sum of Logarithms
\(\ds \) \(=\) \(\ds \frac 1 a \arctan \frac x a + C\) Arctangent Logarithmic Formulation and substituting back $t \to \dfrac x a$

$\blacksquare$


Also see


Sources