<p>If&nbsp;\(\mathrm{F}\prime(x)=\mathrm{f}(x)\), then a linear inner function contributes a constant scale factor:</p>
<p style="text-align: center">\(\int\mathrm{f}(ax+b)\,\mathrm{d}x=\dfrac{1}{a}\mathrm{F}(ax+b)+C\qquad(a\ne0)\)</p>
<div class="content-block content-block--table"><div class="table-container"><table style="width: 100%;"><thead><tr><th style="text-align: center; --cell-justify: center"><strong>Integrand</strong></th><th style="text-align: center; --cell-justify: center"><strong>Antiderivative</strong></th></tr></thead><tbody><tr><td style="text-align: center; --cell-justify: center">\((ax+b)^n\)</td><td style="text-align: center; --cell-justify: center">\(\dfrac{(ax+b)^{n+1}}{a(n+1)}+C\)</td></tr><tr><td style="text-align: center; --cell-justify: center">\(\dfrac{1}{ax+b}\)</td><td style="text-align: center; --cell-justify: center">\(\dfrac{1}{a}\ln|ax+b|+C\)</td></tr><tr><td style="text-align: center; --cell-justify: center">\(e^{ax+b}\)</td><td style="text-align: center; --cell-justify: center">\(\dfrac{1}{a} e^{ax+b}+C\)</td></tr><tr><td style="text-align: center; --cell-justify: center">\(\sin(ax+b)\)</td><td style="text-align: center; --cell-justify: center">\(-\dfrac{1}{a}\cos(ax+b)+C\)</td></tr><tr><td style="text-align: center; --cell-justify: center">\(\cos(ax+b)\)</td><td style="text-align: center; --cell-justify: center">\(\dfrac{1}{a}\sin(ax+b)+C\)</td></tr><tr><td style="text-align: center; --cell-justify: center">\(\sec^2(ax+b)\)</td><td style="text-align: center; --cell-justify: center">\(\dfrac{1}{a}\tan(ax+b)+C\)</td></tr><tr><td style="text-align: center; --cell-justify: center">\(\operatorname{cosec}^2(ax+b)\)</td><td style="text-align: center; --cell-justify: center">\(-\dfrac{1}{a}\cot(ax+b)+C\)</td></tr></tbody></table></div></div>
<p>In the power rule, \(n\in\mathbb Q\) and \(n\ne-1\); choose a real interval on which the expression is defined. The reciprocal form requires \(ax+b\ne0\). Trigonometric denominators must also be non-zero.</p>
<p>If&nbsp;\(\mathrm{F}\prime(x)=2x\) and \(\mathrm{F}(1)=5\), then&nbsp;\(\mathrm{F}(x)=x^2+C\), \(C=4\), and \(\mathrm{F}(x)=x^2+4\).</p>
<p>Dividing by a constant works only for a linear inner function. For&nbsp;\(\int\dfrac{1}{3x^2-1}\,\mathrm{d}x\), the inner derivative is \(6x\), so use another technique.</p>

If \(\mathrm{F}\prime(x)=\mathrm{f}(x)\), then a linear inner function contributes a constant scale factor:

\(\int\mathrm{f}(ax+b)\,\mathrm{d}x=\dfrac{1}{a}\mathrm{F}(ax+b)+C\qquad(a\ne0)\)

In the power rule, \(n\in\mathbb Q\) and \(n\ne-1\); choose a real interval on which the expression is defined. The reciprocal form requires \(ax+b\ne0\). Trigonometric denominators must also be non-zero.

If \(\mathrm{F}\prime(x)=2x\) and \(\mathrm{F}(1)=5\), then \(\mathrm{F}(x)=x^2+C\), \(C=4\), and \(\mathrm{F}(x)=x^2+4\).

Dividing by a constant works only for a linear inner function. For \(\int\dfrac{1}{3x^2-1}\,\mathrm{d}x\), the inner derivative is \(6x\), so use another technique.

Standard Integrals and Linear Composites

Standard Integrals and Linear Composites

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Solution:

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Integrand	Antiderivative
\((ax+b)^n\)	\(\dfrac{(ax+b)^{n+1}}{a(n+1)}+C\)
\(\dfrac{1}{ax+b}\)	\(\dfrac{1}{a}\ln\|ax+b\|+C\)
\(e^{ax+b}\)	\(\dfrac{1}{a} e^{ax+b}+C\)
\(\sin(ax+b)\)	\(-\dfrac{1}{a}\cos(ax+b)+C\)
\(\cos(ax+b)\)	\(\dfrac{1}{a}\sin(ax+b)+C\)
\(\sec^2(ax+b)\)	\(\dfrac{1}{a}\tan(ax+b)+C\)
\(\operatorname{cosec}^2(ax+b)\)	\(-\dfrac{1}{a}\cot(ax+b)+C\)