5.10 Indefinite integrals and substitutionIB Maths: Analysis and Approaches HL: Revision notes

Integration reverses differentiation. The indefinite integral always includes a constant of integration $C$: $\int x^{n}\,dx=\frac{x^{n+1}}{n+1}+C\ (n\in\mathbb{Q},\ n\neq-1),\qquad\int\frac{1}{x}\,dx=\ln|x|+C,$ $\int\sin x\,dx=-\cos x+C,\quad\int\cos x\,dx=\sin x+C,\quad\int e^{x}\,dx=e^{x}+C.$ These are in the formula booklet. Rewrite roots and fractions as powers first: $\int3\sqrt{x}\,dx=\int3x^{1/2}\,dx=2x^{3/2}+C$.

If the inside function is linear, $ax+b$, integrate as normal and divide by $a$: $\int f(ax+b)\,dx=\frac{1}{a}F(ax+b)+C\quad\text{where }F'=f.$

• $\int\cos(2x+3)\,dx=\frac12\sin(2x+3)+C$
• $\int e^{1-2x}\,dx=-\frac12e^{1-2x}+C$
• $\int\frac{4}{2x+1}\,dx=2\ln|2x+1|+C$
• $\int(3x-1)^{4}\,dx=\frac{(3x-1)^{5}}{15}+C$

When the integrand is a multiple of $g'(x)f(g(x))$, the answer is a multiple of $F(g(x))$. Guess the form, differentiate it, and adjust the constant:

$\int2x(x^{2}+1)^{4}\,dx$: try $(x^{2}+1)^{5}$, whose derivative is $10x(x^{2}+1)^{4}$, five times too big. So the answer is $\frac{(x^{2}+1)^{5}}{5}+C$.

A useful special case: $\int\frac{g'(x)}{g(x)}\,dx=\ln|g(x)|+C$, e.g. $\int\frac{x}{x^{2}+5}\,dx=\frac12\ln(x^{2}+5)+C$.

For $\int k\,g'(x)f(g(x))\,dx$:

1. Let $u=g(x)$ and find $\frac{du}{dx}=g'(x)$, so $du=g'(x)\,dx$.
2. Replace everything, including $dx$, so the integral is in $u$ only.
3. Integrate with respect to $u$.
4. Substitute back to give the answer in $x$.

Example: $\int4x\sin(x^{2})\,dx$ with $u=x^{2}$, $du=2x\,dx$: $\int2\sin u\,du=-2\cos u+C=-2\cos(x^{2})+C$.

Example: $\int\frac{\ln x}{x}\,dx$ with $u=\ln x$, $du=\frac1x\,dx$: $\int u\,du=\frac12(\ln x)^{2}+C$.

If you know one point on the curve (or one value of a quantity), substitute it to find $C$. For $f'(x)=\frac{\ln x}{x}$ with $f(e)=1$: $f(x)=\frac12(\ln x)^{2}+C$, $\frac12+C=1$, $C=\frac12$.

In context, the starting value fixes the constant: a reservoir filling at $R(t)=24te^{-0.2t^{2}}$ with $V(0)=500$ has $V(t)=560-60e^{-0.2t^{2}}$, which tends to 560 as $t\to\infty$.