Primitive of Reciprocal of x squared plus a squared/Arctangent Form/Corollary 2

Corollary to Primitive of Reciprocal of x squared plus a squared

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

where $a$ and $b$ are non-zero constants.

Proof

Let $z = b x$.

Then:

$\dfrac {\d x} {\d z} = \dfrac 1 b$

Hence:

\(\ds \int \frac {\d x} {a^2 + b^2 x^2}\)

\(=\)

\(\ds \int \dfrac 1 b \frac {\d z} {a^2 + z^2}\)

Integration by Substitution

\(\ds \)

\(=\)

\(\ds \dfrac 1 b \cdot \frac 1 a \arctan \frac z a + C\)

Primitive of $\dfrac 1 {a^2 + z^2}$: Arctangent Form

\(\ds \)

\(=\)

\(\ds \frac 1 {a b} \arctan \frac {b x} a + C\)

subtituting for $z$ and simplifying

$\blacksquare$

Sources

• 1964: Milton Abramowitz and Irene A. Stegun: Handbook of Mathematical Functions ... (previous) ... (next): $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Rational Algebraic Functions: $3.3.21$