Integrals of the Form $\int f(ax+b)\,dx$

The pattern that motivates this section. Every standard antiderivative on the table in §11.3 was built for a bare variable $x$. What happens when the argument is not $x$ itself but a linear expression $ax+b$ — for instance $\sin(2x+4)$ or $(4x+5)^6$ instead of $\sin x$ or $x^6$?

Constants added/subtracted change nothing. From $\dfrac{d}{dx}\!\left[\dfrac{(x-a)^{10}}{10}\right]=(x-a)^9$ we recover $\displaystyle\int (x-a)^9\,dx=\dfrac{(x-a)^{10}}{10}+c$; likewise $\dfrac{d}{dx}[\sin(x+k)]=\cos(x+k)$ gives $\displaystyle\int\cos(x+k)\,dx=\sin(x+k)+c$. So whenever a constant is merely added to or subtracted from $x$ inside the argument, the fundamental antiderivative formula carries straight over unchanged — no correction factor needed.

A coefficient of $x$ DOES introduce a correction factor. But when $x$ is scaled — the argument is $ax+b$ rather than just $x+b$ — an extra factor appears. Differentiating $\dfrac1l e^{lx+m}$ by the chain rule gives $e^{lx+m}$, so $\displaystyle\int e^{lx+m}\,dx=\dfrac1l e^{lx+m}+c$; similarly $\dfrac{d}{dx}\!\left[\dfrac1a\sin(ax+b)\right]=\cos(ax+b)$ gives $\displaystyle\int \cos(ax+b)\,dx=\dfrac1a\sin(ax+b)+c$. In both cases the extra factor is exactly $\dfrac1a$ — the reciprocal of the coefficient of $x$ — because the chain rule multiplies the derivative of the outer function by the derivative of the inner linear argument, which is the constant $a$; integrating undoes that multiplication by dividing by $a$.

Applying the rule across function families.

• Powers of a linear argument. $\displaystyle\int (4x+5)^6\,dx = \frac14\cdot\frac{(4x+5)^7}{7}+c=\frac{(4x+5)^7}{28}+c$. A negative or fractional exponent works identically: $\displaystyle\int \sqrt{15-2x}\,dx=\int(15-2x)^{1/2}dx=\left(\frac{1}{-2}\right)\frac{(15-2x)^{3/2}}{3/2}+c=-\frac{(15-2x)^{3/2}}{3}+c$, and $\displaystyle\int \frac{dx}{(3x+7)^4}=\int(3x+7)^{-4}dx=\frac13\cdot\frac{(3x+7)^{-3}}{-3}+c=-\frac{1}{9(3x+7)^3}+c$.
• Trigonometric arguments. $\displaystyle\int \sin(2x+4)\,dx=\frac12\big(-\cos(2x+4)\big)+c=-\frac12\cos(2x+4)+c$; $\displaystyle\int \sec^2(3+4x)\,dx=\frac14\tan(3+4x)+c$; and $\displaystyle\int \text{cosec}(ax+b)\cot(ax+b)\,dx=-\frac1a\,\text{cosec}(ax+b)+c$.
• Exponential arguments. $\displaystyle\int e^{3x}\,dx=\frac13e^{3x}+c$, and with a negative coefficient $\displaystyle\int e^{5-4x}\,dx=-\frac14e^{5-4x}+c$ (the sign of $a$ simply carries through the $\tfrac1a$ factor).
• Reciprocal-linear ($\log$) arguments. $\displaystyle\int \frac{dx}{3x-2}=\frac13\log|3x-2|+c$, and $\displaystyle\int \frac{dx}{5-4x}=-\frac14\log|5-4x|+c$. …