Integrating the reciprocal function

The integral of 1/x as ln|x|, the standard forms for 1/(ax+b) and f'(x)/f(x), integrals over negative intervals, and finding the integral of ln x from a given derivative.

Standard forms

Reciprocal integrals

$$\int \frac1x\,dx = \ln|x| + C \qquad \int \frac{1}{ax + b}\,dx = \frac1a\ln|ax + b| + C \qquad \int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C$$

Example 1●

Evaluate (a) $\displaystyle\int_1^e \frac3x\,dx$ and (b) $\displaystyle\int_0^2 \frac{1}{2x + 1}\,dx$.

$3\Big[\ln x\Big]_1^e = 3(1 - 0) = 3$

$\tfrac12\Big[\ln(2x + 1)\Big]_0^2 = \tfrac12(\ln 5 - \ln 1) = \tfrac12\ln 5$

Example 2: negative values●●

Evaluate $\displaystyle\int_{-3}^{-1} \frac1x\,dx$.

$\Big[\ln|x|\Big]_{-3}^{-1} = \ln 1 - \ln 3 = -\ln 3$. The answer is negative because the curve is below the axis there.

The numerator is the derivative of the denominator

Example 3●●

Evaluate $\displaystyle\int_0^1 \frac{x}{x^2 + 1}\,dx$.

$\tfrac12\displaystyle\int_0^1 \frac{2x}{x^2 + 1}\,dx = \tfrac12\Big[\ln(x^2 + 1)\Big]_0^1 = \tfrac12\ln 2$

Example 4: divide first●●

Evaluate $\displaystyle\int_1^2 \frac{x + 2}{x}\,dx$.

$\dfrac{x + 2}{x} = 1 + \dfrac2x$, so the integral is $\Big[x + 2\ln x\Big]_1^2 = 1 + 2\ln 2$.

Integrals from a given derivative

Example 5●●●

Differentiate $x\ln x$, and hence find $\displaystyle\int \ln x\,dx$.

$\dfrac{d}{dx}x\ln x = \ln x + 1$, so $\displaystyle\int (\ln x + 1)\,dx = x\ln x + C$.

$\displaystyle\int \ln x\,dx = x\ln x - x + C$

Flashcards

$displaystyleint rac1x,dx$

$\ln|x| + C$

$displaystyleint rac{f'(x)}{f(x)},dx$

$\ln|f(x)| + C$

$displaystyleint ln x,dx$

$x\ln x - x + C$

Practice

$\displaystyle\int \frac1x\,dx = $
Evaluate $\displaystyle\int_1^e \frac3x\,dx$.
Evaluate $\displaystyle\int_0^2 \frac{1}{2x + 1}\,dx$. Type $\ln 5$ as ln5.
$\displaystyle\int \frac{2x}{x^2 + 1}\,dx = $
Evaluate $\displaystyle\int_{-3}^{-1} \frac1x\,dx$.
Evaluate $\displaystyle\int_1^2 \frac{x + 2}{x}\,dx$.
$\displaystyle\int \ln x\,dx = $
Evaluate $\displaystyle\int_1^e \ln x\,dx$.
$\displaystyle\int \frac{1}{3x}\,dx = $