Primitive of Reciprocal

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Theorem

$\ds \int \frac {\d x} x = \ln \size x + C$

for $x \ne 0$.


Corollary 1

$\ds \int \frac {\d x} x = \ln x + C$

for $x > 0$.


Corollary 2

$\dfrac {\d} {\d x} \ln \size x = \dfrac 1 x$

for $x \ne 0$.


Proof

Suppose $x > 0$.

Then:

$\ln \size x = \ln x$

The result follows from Derivative of Natural Logarithm Function and the definition of primitive.


Suppose $x < 0$.

Then:

\(\ds \dfrac \d {\d x} \ln \size x\) \(=\) \(\ds \dfrac \d {\d x} \map \ln {-x}\) Definition of Absolute Value
\(\ds \) \(=\) \(\ds \frac 1 {-x} \cdot -1\) Chain Rule for Derivatives and Derivative of Natural Logarithm Function, as $-x > 0$
\(\ds \) \(=\) \(\ds \frac 1 x\)

and the result again follows from the definition of the primitive.

$\blacksquare$


Also presented as

Some sources gloss over the case where $x < 0$ and merely present this result as:

$\ds \int \frac {\d x} x = \ln x + C$


Examples

Integral of $\dfrac 1 {x - 5}$ from $2$ to $3$

$\ds \int_2^3 \dfrac {\d x} {x - 5} = \ln \dfrac 2 3$


Sources

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