Integration by Parts

Formula¶

\[ \int u {dv \over{dx}} \: dx = uv - \int {du \over {dx}}v \: dx \]

Ex.

\[ ∫xe^xdx \]

if u = x and \({dv \over dx} = e^xdx\)

Then: \(v = ∫e^x\) and \({du\over dx} = u'\)

L.I.A.T.E

L: logs (\(\log{x}\), \(\log{2x}\), \(\log_{10}{x}\))

I: inverse trig (\(\arctan{x}\), \(\arcsin{2x}\))

A: Algebraic (\(x^2\), \((x-1)\), \(x^6\), \(\sqrt{x}\))

T: Trig Functions (\(\sin{x}\), \(\cos{3x}\))

E: Exponentials (\(ex\), \(e − 3x\), \(2ex + 4\))

Where u is the first type you come across in the expression, and v is the second.