Standard Integrals and Linear Replacements
Standard Integrals and Linear Replacements
When \(x\) is replaced by a linear function \(ax+b\), where \(a\ne0\), the chain rule contributes the constant factor \(a\). Divide the antiderivative by \(a\).
From a base integral to a linear replacement
| Base integral | Linear replacement | Conditions |
|---|---|---|
| \(\displaystyle\int x^n\,\mathrm{d}x=\dfrac{x^{n+1}}{n+1}+C\) | \(\displaystyle\int (ax+b)^n\,\mathrm{d}x=\dfrac{(ax+b)^{n+1}}{a(n+1)}+C\) | \(n\ne-1,\ a\ne0\) |
| \(\displaystyle\int e^x\,\mathrm{d}x=e^x+C\) | \(\displaystyle\int e^{ax+b}\,\mathrm{d}x=\dfrac{e^{ax+b}}{a}+C\) | \(a\ne0\) |
| \(\displaystyle\int \dfrac{1}{x}\,\mathrm{d}x=\ln|x|+C\) | \(\displaystyle\int \dfrac{1}{ax+b}\,\mathrm{d}x=\dfrac{1}{a}\ln|ax+b|+C\) | \(a\ne0,\ x\ne0,\ ax+b\ne0\) |
| \(\displaystyle\int \cos x\,\mathrm{d}x=\sin x+C\) | \(\displaystyle\int \cos(ax+b)\,\mathrm{d}x=\dfrac{\sin(ax+b)}{a}+C\) | \(a\ne0\) |
| \(\displaystyle\int \sin x\,\mathrm{d}x=-\cos x+C\) | \(\displaystyle\int \sin(ax+b)\,\mathrm{d}x=-\dfrac{\cos(ax+b)}{a}+C\) | \(a\ne0\) |
| \(\displaystyle\int \sec^2x\,\mathrm{d}x=\tan x+C\) | \(\displaystyle\int \sec^2(ax+b)\,\mathrm{d}x=\dfrac{\tan(ax+b)}{a}+C\) | \(a\ne0\) |
| \(\displaystyle\int \mathrm{cosec}^2x\,\mathrm{d}x=-\cot x+C\) | \(\displaystyle\int \mathrm{cosec}^2(ax+b)\,\mathrm{d}x=-\dfrac{\cot(ax+b)}{a}+C\) | \(a\ne0\) |
\[\frac{\mathrm{d}}{\mathrm{d}x}(ax+b)=a\]
For example, \(\displaystyle\int\dfrac{1}{3x-1}\,\mathrm{d}x=\dfrac13\ln|3x-1|+C\).
The same shortcut does not apply directly to \(\displaystyle\int\dfrac{1}{3x^2-1}\,\mathrm{d}x\), because the denominator is not a linear replacement of \(x\) and its derivative is \(6x\), not a constant.
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