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Lesson 7, Topic 7
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# Integration of Products (Integration by Parts)

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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. $$\scriptsize \int x^2 e^{3x} dx$$

error: