Theory

Theory 1

The method of integration by parts (abbreviated IBP) is applicable when the integrand is a product for which one factor is easily integrated while the other becomes simpler when differentiated.

Integration by parts

Suppose the integral has this format, for some functions $u$ and $v$:

$$\int u\cdot v^{\prime }dx$$

Then the rule says we may convert the integral like this:

$$\int u\cdot v^{\prime }dx\gg \gg u\cdot v-\int u^{\prime }\cdot vdx$$

This technique comes from the product rule for derivatives:

$$(u\cdot v{)}^{\prime }=u^{\prime }\cdot v+u\cdot v^{\prime }$$

Now, if we integrate both sides of this equation, we find:

$$u\cdot v=\int u^{\prime }\cdot vdx+\int u\cdot v^{\prime }dx$$

and the IBP rule follows by algebra.

Extra - Full explanation of integration by parts

(1) Product rule for derivatives:

$$(u\cdot v{)}^{\prime }=u^{\prime }\cdot v+u\cdot v^{\prime }$$

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(2) Integrate both sides:

$$\begin{array}{cccc}& (u\cdot v{)}^{\prime }=u^{\prime }\cdot v+u\cdot v^{\prime }& \gg \gg \int (u\cdot v{)}^{\prime }dx=\int u^{\prime }\cdot vdx+\int u\cdot v^{\prime }dx& \\ & & & \\ & & \gg \gg u\cdot v=\int u^{\prime }\cdot vdx+\int u\cdot v^{\prime }dx& \text{(FTC)}\\ & & & \\ & & \gg \gg \int u\cdot v^{\prime }dx=u\cdot v-\int u^{\prime }\cdot v& \text{(Rearrange)}\end{array}$$

Definite IBP

Definite version of FTC:

$$\begin{array}{c}{\int }_a^b(u\cdot v{)}^{\prime }dx=u\cdot v{|}_a^b\\ \\ \gg \gg {\int }_a^{b}u\cdot v^{\prime }dx=u\cdot v{|}_a^b-{\int }_a^b{u}^{\prime }\cdot v\end{array}$$

Choosing factors well

IBP is symmetrical. How do we know which factor to choose for $u$ and which for $v$?

Here is a trick: the acronym “LIATE” spells out the order of choices – to the left for $u$ and to the right for $v$:

$$\begin{array}{c}\text{LIATE}:\\ \\ u\leftarrow \text{Logarithmic – Inverse\_trig – Algebraic – Trig – Exponential}\to v\end{array}$$