# High School Calculus/Integration by Parts

### Integration by Parts

When integrating products of algebraic and transcendental functions use integration by parts.

$\int u dv = uv - \int v du$

to determine which on you should choose for $u$ use the anagram I.L.A.T.E.

I.L.A.T.E. stands for Inverse, Logarithmic, Algebraic, Trigonometry, Exponential

This anagram is purely a guideline that will help you with picking the u and dv

Ex. 1

$\int x \ln x \operatorname {d}x$

In order to integrate this we must use integration by parts

Using I.L.A.T.E. we choose $\ln x$ to be equal to u

and $\operatorname {d}v = x \operatorname {d}x$

We also need $\operatorname {d}u$ and $v$

$\operatorname {d}u = \left (\frac {1}{x} \right) \operatorname {d}x$

$\int \operatorname {d}v = \int x \operatorname {d}x$

$v = \left (\frac {1}{2} \right) x^2$

Using the integration by parts theorem we get

$\ln (x) x - \int \left (\frac {1}{2} \right) x^2 \left (\frac {1}{x} \right) \operatorname {d}x$

$\ln (x) x - \left (\frac {1}{2} \right) \int \left (\frac {x}{x^2} \right) \operatorname {d}x$

$\ln (x) x - \left (\frac {1}{2} \right) \int x \operatorname {d}x$

$\ln (x) x - \left (\frac {1}{2} \right) \left (\frac {1}{2} \right) x^2$

$\ln (x) x - \left (\frac {x^2}{4} \right)$

It's important to be able identify what you're going to use for u

### Other Problems to Work On

Ex. 2

$\int \cos (x) x^2 \operatorname {d}x$

Ex. 3

$\int \arcsec (x) \left (\frac {1}{x} \right) \operatorname {d}x$