Primitive of Reciprocal of Root of x squared plus a squared/Logarithm Form/Proof 2

Theorem

$\ds \int \frac {\d x} {\sqrt {x^2 + a^2} } = \map \ln {x + \sqrt {x^2 + a^2} } + C$

Proof

Let $y^2 = a^2 + x^2$.

Then:

\(\ds 2 y \frac {\d y} {\d x}\)

\(=\)

\(\ds 2 x\)

Power Rule for Derivatives, Chain Rule for Derivatives

\(\ds \leadsto \ \ \)

\(\ds y \frac {\d y} {\d x}\)

\(=\)

\(\ds x\)

simplification

\(\ds \leadsto \ \ \)

\(\ds \frac {\d y} x\)

\(=\)

\(\ds \frac {\d x} y\)

\(\ds \)

\(=\)

\(\ds \frac {\d x + \d y} {x + y}\)

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\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds \int \frac {\d x} y\)

substituting for $y$

\(\ds \)

\(=\)

\(\ds \frac {\d x + \d y} {x + y}\)

from above

\(\ds \)

\(=\)

\(\ds \ln \size {x + y} + C\)

Primitive of Function under its Derivative

\(\ds \)

\(=\)

\(\ds \ln \size {x + \sqrt {x^2 + a^2} } + C\)

substituting back

\(\ds \)

\(=\)

\(\ds \map \ln {x + \sqrt {x^2 + a^2} } + C\)

argument of $\ln$ always positive

$\blacksquare$

Sources

• 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Integration: Algebraic Integration: $\text {IV}$.