Remember the square root of a negative number gives the imaginary number, i.

A complex number consists of a real part and an imaginary part.

Example,

$5x^2 - 6x + 5$
$x=\frac{6\pm\sqrt{36-100}}{10}$
$x=\frac{6\pm\sqrt{-64}}{10}$
$x=\frac{6\pm 8i}{10}$
$x=0.6+0.8i \text{ and } 0.6-0.8i$

A complex number in Cartesian form: $z = a + bi$ can be expressed in polar form: $z = r(\cos{\theta} + i\sin{\theta})$ using the conversion:

$r^2 = a^2 + b^2$

$\theta = \tan^{-1} \left(\frac{b}{a}\right)$

General rules for partial fractions:

$\frac{P(x)}{(ax+b)(cx+d)}=\frac{A}{ax+b}+\frac{B}{cx+d}$
$\frac{P(x)}{(ax+b)^2}=\frac{A}{ax+b}+\frac{B}{(ax+b)^2}$
$\frac{P(x)}{(ax^2+bx+c)(dx+e)}=\frac{Ax+B}{ax^2+bx+c}+\frac{C}{dx+e}$

General rules for partial fractions:

$\frac{P(x)}{(ax+b)(cx+d)}=\frac{A}{ax+b}+\frac{B}{cx+d}$
$\frac{P(x)}{(ax+b)^2}=\frac{A}{ax+b}+\frac{B}{(ax+b)^2}$
$\frac{P(x)}{(ax^2+bx+c)(dx+e)}=\frac{Ax+B}{ax^2+bx+c}+\frac{C}{dx+e}$

General rules for partial fractions:

$\frac{P(x)}{(ax+b)(cx+d)}=\frac{A}{ax+b}+\frac{B}{cx+d}$
$\frac{P(x)}{(ax+b)^2}=\frac{A}{ax+b}+\frac{B}{(ax+b)^2}$
$\frac{P(x)}{(ax^2+bx+c)(dx+e)}=\frac{Ax+B}{ax^2+bx+c}+\frac{C}{dx+e}$

Example, 

$\displaystyle \int 2xe^x\,dx$
$u = 2x \rightarrow \frac{du}{dx}=2$
$\frac{dv}{dx}=e^x \rightarrow v = \int e^x.dx = e^x$
$\int 2xe^x.dx = (2x)\left(e^x\right) - \int \left(e^x \right) \left(2\right).dx$
$=2xe^x - 2\int e^x.dx$
$=2xe^x - 2e^x + C$
$=2e^x(x-1) + C$

The tabular method could also be used. In this approach, one function is differentiated until it reaches zero. The other is integrated the corresponding number of times. An alternative sign approach is then used to multiple down the table and form the answer. Please see the notes from AMR129 for an example of this. 

Example, 

$\displaystyle \int 2xe^x\,dx$
$u = 2x \rightarrow \frac{du}{dx}=2$
$\frac{dv}{dx}=e^x \rightarrow v = \int e^x.dx = e^x$
$\int 2xe^x.dx = (2x)\left(e^x\right) - \int \left(e^x \right) \left(2\right).dx$
$=2xe^x - 2\int e^x.dx$
$=2xe^x - 2e^x + C$
$=2e^x(x-1) + C$

The tabular method could also be used. In this approach, one function is differentiated until it reaches zero. The other is integrated the corresponding number of times. An alternative sign approach is then used to multiple down the table and form the answer. Please see the notes from AMR129 for an example of this.