Definition
$$ F(x)=\int f(x) \, \text{d}x \quad;\quad \frac{\text{d}F}{\text{d}x}=f(x) $$
$$ \begin{aligned} A&=\int_{a}^{b} f(x) \textrm{d} x \\&= \lim_{\delta x\rightarrow 0} \sum_{x=a}^b f(x) \delta x \\&= [ F(x) ]_a^b = F(b)-F(a) \end{aligned} $$
Properties:
$$ \begin{aligned} \int f(x){\color{Orange}+}g(x) \textrm{d} x&= \int f(x) \textrm{d} x\,{\color{Orange}+}\int g(x) \textrm{d} x \\ \int {\color{Orange}c} f(x) \textrm{d} x&={\color{Orange}c} \int f(x) \textrm{d} x \\ \int_{\color{Orange}a}^{\color{Orange}b} f(x) \textrm{d} x &={\color{Orange}-}\int_{\color{Orange}b}^{\color{Orange}a} f(x) \textrm{d} x \\ \int_{a}^{b} f(x) \textrm{d} x&=\int_{a}^{\color{Orange}c} f(x) \textrm{d} x \,{\color{Orange}+}\int_{\color{Orange}c}^{b} f(x) \textrm{d} x \\ \int_{-a}^{a} f(x) \textrm{d} x &={\color{Orange}0}; \quad \textrm{for an} {\color{Orange}\text{ odd }} \text{function }f(x) \\ \int_{\color{Orange}-a}^{a} f(x) \textrm{d} x &={\color{Orange}2} \int_{\color{Orange}0}^{a} f(x) \textrm{d} x; \quad \textrm{for an} {\color{Orange}\text{ even }} \text{function }f(x) \end{aligned} $$
| $f(x)$ | $F(x)=\int f(x) \, \text{d} x$ |
|---|---|
| $x^n$ | $\frac{x^{n+1}}{n+1}+c\, ; \; n\ne -1$ |
| $\sin{x}$ | $-\cos x + c$ |
| $\cos x$ | $\sin x + c$ |
| $e^x$ | $e^x + c$ |
| $\frac{1}{x}$ | $\ln (x) + c$ for $x>0$ |
| $\sinh x$ | $\cosh x + c$ |
| $\cosh x$ | $\sinh x + c$ |
$$ \begin{aligned} \int \frac{f'(x)}{f(x)} \textrm{d} x &= \ln | f(x)|+c \\ \int f(x) f'(x) \textrm{d} x &= \frac{1}{2} f^2(x)+c \\ \int \frac{f'(x)}{\sqrt{f(x)}} \textrm{d} x &= 2(f(x))^{1/2}+c \\ \int \frac{1}{x} \mathrm{d}x &= \ln |f(x)|+c \end{aligned} $$
$$ \int_{a}^{b} f(x) \textrm{d} x =\lim\limits_{\epsilon \to 0} \int_{a}^{b-\epsilon} f(x) \textrm{d} x \; ; \; \int_{a}^{b} f(x) \textrm{d} x =\int_{a}^{c} f(x) \textrm{d} x+\int_{c}^{b} f(x) \textrm{d} x $$
$$ \int_1^\infty x^{-2}\textrm{d}x = \lim\limits_{\epsilon \to \infty} \int_1^\epsilon x^{-2}\textrm{d}x = \lim\limits_{\epsilon \to \infty}(1-\frac{1}{\epsilon})=1 $$
| $f(x)$ | $F(x)=\int f(x) \text\,{d}x$ | Substitution |
|---|---|---|
| $\frac{1}{\sqrt{a^2-x^2}}$ | $\arcsin(\frac{x}{a})+ c$ | $x=a\sin\theta$ |
| $\frac{1}{\sqrt{x^2-a^2}}$ | $\text{arccosh} (\frac{x}{a}) +c$ | $x=a\cosh u$ |
| $\frac{1}{\sqrt{x^2+a^2}}$ | $\text{arcsinh} (\frac{x}{a})+ c$ | $x=a\sinh u$ |
| $\frac{1}{x^2+a^2}$ | $\frac{1}{a}\text{arctan}(\frac{x}{a})+ c$ | $x=a\tan\theta$ |
Example: Evaluate the integral: $I=\int_{0}^{4} \frac{1}{1+\sqrt{x}} \textrm{d} x$
We can take $x=u^2$ → $\text{d}x=\frac{\text{d}x}{\text{d}u}\text{d}u=2u\text{d}u$
lim are: $\int^4_0f_x\,\text{d}x=\int^{\sqrt{4}}_{\sqrt{0}}f_u\,\text{d}u=\int^2_0f_u\,\text{d}u$
$$ \begin{aligned} I&=\int_{0}^{2} \frac{2u}{1+u} \textrm{d} u =\int_{0}^{2} \frac{2u+2 -2}{1+u} \textrm{d} u \\ &=\int_{0}^{2} 2\frac{u+1}{1+u} -\frac{2}{1+u} \textrm{d} u\\&=\int_{0}^{2} 2-\frac{2}{1+u} \textrm{d} u \\ &=\left[2 u-2 \ln(1+u) \right]_0^{2}\\&=(4-2 \ln3) -(0-0) \\ I&=4-2\ln3 \end{aligned} $$
$$ \begin{aligned} I&=\int \frac{\overbrace{p x+q}^{\frac{1}{2} p(2x +a)+(q- \frac{1}{2} pa )}}{x^2+a x+b} \textrm{d} x\\ I&=\frac{1}{2}p \overbrace{\int \frac{2 x+a}{x^2+a x+b} \textrm{d} x}^{\ln|x^2+a x+b|}+ (q-\frac{1}{2}pa ) \overbrace{\int \frac{\textrm{d} x}{\underbrace{x^2+a x+b}_{(x+\frac{a}{2})^2+b-(\frac{a}{2})^2}}}^{\frac{1}{\sqrt{b- (\frac{a}{2})^2}} \arctan (\frac{x+a/2}{\sqrt{b- ( \frac{a}{2})^2}})} \\ I&=\frac{1}{2}p \ln(|x^2+a x+b|)+ (q-\frac{1}{2}pa )\frac{1}{\sqrt{b- \left(\frac{a}{2}\right)^2}} \arctan\left(\frac{x+a/2}{\sqrt{b-(a/2)^2}}\right)+c \end{aligned} $$
Example:
$$ \text{Solve: }\int\frac{3x^2+2x+1}{2x^3+5x^2+4x+1}\text{d}x \\ I= \int\left[\Bigl\{ \frac{3x^2+2x+1}{\underbrace{2x^3+5x^2+4x+1}{(2x+1)(x+1)^2}}\Bigl\}\Rightarrow \frac{A}{2x+1}+\underbrace{\frac{B}{x+1}+\frac{C}{(x+1)^2}}{\text{for repeated factors}}\right]\text{d}x \\
$$
Lets try and find $A\;B\;C$
$$ \overbrace{\left[ \frac{3x^2+2x+1}{2x^3+5x^2+4x+1}=\frac{A}{2x+1}+\frac{B}{x+1}+\frac{C}{(x+1)^2}\right]}^{\text{we know the following}} \cdot \overbrace{(2x+1)(x+1)^2}^{\substack{\text{attempt to eliminate} \\ \text{terms}}} \\ \Rightarrow 3x^2+2x+1=A(x+1)^2+B(2x+1)(x+1)+C(2x+1) \\ \begin{aligned} x=-1 &\rightarrow 2=-C \rightarrow C=-2 \\ x=-\frac{1}{2} &\rightarrow0.75=\frac{1}{4}A \rightarrow A=3 \\ x=0 &\rightarrow 1= A+B+C \rightarrow B=1-A-C=1-3+2=0 \end{aligned}
$$
Now we can simply solve