We often need to integrate a product where either function is not the derivative of the other.
If u and v are functions of x, then
\( \frac{d(uv)}{dx} = \scriptsize u \normalsize\frac{dv}{dx} + \scriptsize v \normalsize\frac{du}{dx} \)Integrating both sides with respect to x
uv = \(\int \scriptsize u \normalsize\frac{dv}{dx} \scriptsize dx + \int \scriptsize v \normalsize\frac{du}{dx} \scriptsize dx \)
uv = \(\int \scriptsize udv + \normalsize \int \scriptsize vdu \)
\(\int \scriptsize udv = uv\; – \normalsize \int \scriptsize vdu \)Guides for selecting u and dv in a given function.
Select u in this order below:
L: logarithmic function
I: inverse trigonometric function
A: algebraic function
T: trigonometric function
E: exponential function
Example:
Evaluate \(\scriptsize \int x^3 \ln x dx \)
Solution: \(\int \scriptsize udv = uv\; – \normalsize \int \scriptsize vdu \)
Let u = ln x since it’s a logarithmic function, then dv = \( \scriptsize x^3 dx \)
We obtain \( \frac{du}{dx} = \frac{1}{dx}, \scriptsize du = \normalsize \frac{1}{x} \scriptsize dx\) and
v = \(\scriptsize \int x^3 dx = \normalsize \frac{x^4}{4} \)
Substituting into the formula,
\(\int \scriptsize udv = uv\; – \normalsize \int \scriptsize vdu \) we have
= \( \scriptsize \ln x \normalsize \frac{x^4}{4} \; – \frac{x^4}{4} \frac{1}{x} \scriptsize dx \)
= \( \scriptsize \ln x \normalsize \frac{x^4}{4} \; – \frac{1}{4}\scriptsize \int x^3 dx \)
= \( \scriptsize \ln x \normalsize \frac{x^4}{4} \; – \frac{1}{4}\left( \frac{x^4}{4} \right) \scriptsize + c \)
\(\scriptsize \int x^3 \ln x dx = \normalsize \frac{x^4}{4} \left ( \scriptsize \ln x \; – \normalsize \frac{1}{4} \right)\scriptsize + c \)Exercise
Evaluate the following
a. \(\scriptsize \int x^2 \ln x dx \)
b. \(\scriptsize \int x^2 e^{3x} dx \)
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