Porisma

Integrate as reverse differentiation

Recognise standard antiderivatives and account for a linear inner function.

Integrate as reverse differentiation

If F′(x)=f(x)F'(x)=f(x), then ∫f(x) dx=F(x)+C\int f(x)\,dx=F(x)+C, where C is an arbitrary constant. Antiderivatives are considered on an interval where the expression is defined; additive constants may differ between disconnected intervals. The source linearity rules are

∫[f(x)±g(x)] dx=∫f(x) dx±∫g(x) dx,∫kf(x) dx=k∫f(x) dx(k constant).\begin{aligned} \int[f(x)\pm g(x)]\,dx&=\int f(x)\,dx\pm\int g(x)\,dx,\\ \int kf(x)\,dx&=k\int f(x)\,dx\quad(k\text{ constant}). \end{aligned}

All basic and linear-inner-function rows are retained, with a≠0 in the right column:

Involving xInvolving ax+b, a≠0
∫xn dx=xn+1n+1+C\int x^n\,dx=\frac{x^{n+1}}{n+1}+C, n≠−1∫(ax+b)n dx=(ax+b)n+1(n+1)a+C\int(ax+b)^n\,dx=\frac{(ax+b)^{n+1}}{(n+1)a}+C, n≠−1
∫ex dx=ex+C\int e^x\,dx=e^x+C∫eax+b dx=eax+ba+C\int e^{ax+b}\,dx=\frac{e^{ax+b}}a+C
∫cos⁡x dx=sin⁡x+C\int\cos x\,dx=\sin x+C∫cos⁡(ax+b) dx=sin⁡(ax+b)a+C\int\cos(ax+b)\,dx=\frac{\sin(ax+b)}a+C
∫sin⁡x dx=−cos⁡x+C\int\sin x\,dx=-\cos x+C∫sin⁡(ax+b) dx=−cos⁡(ax+b)a+C\int\sin(ax+b)\,dx=-\frac{\cos(ax+b)}a+C
∫sec⁡2x dx=tan⁡x+C\int\sec^2x\,dx=\tan x+CNo linear-inner counterpart is printed in this row.

Power expressions use their real domains, and trigonometric identities use radians. The n=−1 case is excluded from this printed power formula; it is not obtained by dividing by zero. Example 29 retains all four asks, including its final worked part on the following source page. Continue with Tutorial Question 18.

Try these examples

Open an example in chapter practice. Try it first, then read its solution. Use “Learn this step” to return to this unit.