We use the reverse chain rule to integrate expressions of the form \( \int \ k \frac{f'(x)}{f(x)}\) and \( \int \ kf'(x) (f(x))^n \)

For \( \int \ k \frac{f'(x)}{f(x)}\) we use \( ln|f(x)| \) as our trial solution, then differentiate to find our scale factor.

For \( \int \ kf'(x) (f(x))^n \) we use \( (f(x))^{n+1} \) as our trial solution, then differentiate to find our scale factor.

Examples

Find \( \int \frac{3x}{x^2+4}\ dx \)

\begin{align} Let \quad I & = \int \frac{3x}{x^2+4}\ dx \\ \\ And \quad let \quad f(x)& = x^2+4 \\ \\ Then \quad f'(x) & = 2x \\ \\ So\ we\ know\ \int \frac{2x}{x^2+4} & = ln|x^2+4|\\ \\ (Now\ scale\ by\ & \frac{3}{2}\ in\ order\ to\ produce\\ the\ required\ numerator\,, 3x\,, & \ after\ differentiation ) \\ \\ So \quad I= \frac{3}{2} ln|x^2+4|+c\\ \end{align}

Find \( \int \frac{sinx}{3cosx-5}\ dx \)

\begin{align} Let \quad I & = \int \frac{sinx}{3cosx-5}\ dx \\ \\ And \quad let \quad f(x)& = 3cosx-5 \\ \\ Then \quad f'(x) & = -3sinx \\ \\ So\ we\ know\ \int \frac{-3sinx}{3cosx-5} & = ln|3cosx-5|\\ (Now\ scale\ by\ & -\frac{1}{3}\ in\ order\ to\ produce\\ the\ required\ numerator\,, sinx\,, & \ after\ differentiation ) \\ \\ So \quad I= -\frac{1}{3} ln|3cosx-5|+c\\ \end{align}

Find \( \int 5sin^2x\,cosx \ dx \)

\begin{align} Let \quad I & = \int 5sin^2x\,cosx\ dx \\ \\ And \quad let \quad f(x)& = sin^3x \\ \\ Then \quad f'(x) & = 3sin^2x\,cosx \\ \\ So\ we\ know\ \int 3sin^2x\, cosx & = sin^3x \\ (Now\ scale\ by\ & \frac{5}{3}\ in\ order\ to\ produce\\ the\ required\ coefficient\,, 5\,, & \ after\ differentiation ) \\ \\ So \quad I= \frac{5}{3} sin^3x+c\\ \end{align}

Find \( \int 7x(x^2-4)^3 \ dx \)

\begin{align} Let \quad I & = \int x(x^2-4)^3\ dx \\ \\ And \quad let \quad f(x)& = (x^2-4)^4 \\ \\ Then \quad f'(x) & = 4(x^2-4)^3\, 2x \\ \\ So\ we\ know\ \int\ 8x(x^2-4)^3\ dx & = (x^2-4)^4\ \\ (Now\ scale\ by\ & \frac{7}{8}\ in\ order\ to\ produce\ the \\ required\ coefficient\ of\ x\,, 7\,, & \ after\ differentiation ) \\ \\ So \quad I= \frac{7}{8}(x^2-4)^4+c\\ \end{align}