High School Calculus/Integration by Parts

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Integration by Parts[edit]

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[edit]

Ex. 2

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

Ex. 3

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