Integration by Parts

#Calc-2#Math#Integrals Integration using substitution is the chain rule in reverse.

Similarly, Integration by Parts is the product rule in reverse

We have two functions, u and v, both functions of x The product rule of these two functions would give us $u{v}^{\prime }+v{u}^{\prime }$ , which can be rewritten in terms of integrals as $\int u{v}^{\prime }dx+\int v{u}^{\prime }dx$ These can be further simplified into $\int udv+\int vdu$. Through a bit of algebra we end up with the final formula, $\int udv=uv-\int vdu$ We choose u within our integral based off the following acronym

LIATE §

L - logarithms I - Inverse Trig Functions A - Algebraic Functions T - Trig functions E - Exponential Functions

choosing u based off this will let us end up with an easier integral in the end.

Choose u based off of this, make the rest of the function dv. Then find du and v and plug everything into the equation above.

This rule can be violated, its just a rule of thumb but it usually works

Example §

$\int \ln (x)xdx$ since $\ln (x)$ is a logarithm, which is the highest on our list, we choose that as u

$u=\ln (x),dv=xdx$ $du=\frac{1}{x}dx,v=\frac{x^2}{2}$ don’t bother including +C in the v term, since you’ll add one in at the end anyways.

$\ln (x)\frac{x^2}{2}-\int \frac{x^2}{2}\frac{1}{x}dx$ $\ln (x)\frac{x^2}{2}-\frac{1}{2}\int xdx$ $\frac{1}{2}{x}^2\ln (x)-\frac{1}{4}{x}^2+C$

In a definite integral, the bounds of the integral remain the same, and don’t forget to evaluate the uv part.