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Lesson 7, Topic 6
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# Integration by Partial Fraction

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Expression such as $$\scriptsize \int \frac{7x + 8}{2x^2 + 11x + 5} dx$$do not appear in the list of standard integrals but do occur in many mathematical applications.

Such expressions can be expressed in partial fractions which are simpler in structure.

$$\frac{7x + 8}{2x^2 + 11x + 5} = \frac{A}{x + 5} + \frac{B}{2x + 1}$$

We then proceed to find A and B.

7x + 8 = A(2x +1) + B(x + 5), let x = -5

Then we have,

7(-5) + 8 = A(2(-5) +1) + B( -5 + 5)

-35 + 8 = A( -10 +1)

-27 = -9A

A = 3

Let x = $$-\frac{1}{2}$$

Then $$\scriptsize 7 \left (-\frac{1}{2} \right) + 8 = A \left [2 \left(-\frac{1}{2} \right) + 1 \right] + B(-\frac{1}{2} + 5 )$$

= $$\scriptsize 7 \left (-\frac{1}{2} \right) + 8 = A \left [ -1 + 1 \right] + B(-\frac{1}{2} + 5)$$

= $$\scriptsize – 3 \frac{1}{2} + 8 = B(-\frac{1}{2} + 5)$$

= $$\scriptsize 4 \frac{1}{2} = B 4 \frac{1}{2}$$

$$\frac{9}{2} = \frac{9}{2}\scriptsize B$$ $$\scriptsize B = \frac{9}{2} \; \times \frac{2}{9}$$ $$\therefore \scriptsize B = 1$$

Therefore $$\frac{7x + 8}{2x^2 + 11x + 5} = \frac{3}{x + 5} + \frac{1}{2x + 1}$$

$$\int \frac{7x + 8}{2x^2 + 11x + 5} \scriptsize dx = \normalsize\int\frac{3}{x + 5}\scriptsize dx + \normalsize\int \frac{1}{2x + 1}\scriptsize dx$$

= $$\scriptsize 3 \normalsize \int\frac{1}{x + 5}\scriptsize dx + \normalsize\int \frac{1}{2x + 1}\scriptsize dx$$

We go to integrate the two terms by substitution

$$\scriptsize 3 \normalsize \int\frac{1}{u}\scriptsize dx + \normalsize\int \frac{1}{u}\scriptsize dx$$

$$\scriptsize 3 \normalsize \int \frac{1}{u} \scriptsize dx \rightarrow u = x + 5, \normalsize\frac{du}{dx} \scriptsize = 1 \therefore du = dx$$

$$\scriptsize 3 \normalsize \int\frac{d}{u}\scriptsize$$ = u = 3ln(x + 5)

$$\scriptsize 3 \normalsize \int\frac{1}{u}\scriptsize dx \rightarrow u = 2x + 1, \normalsize\frac{du}{dx} \scriptsize = 2 \therefore dx = \normalsize\frac{du}{2}$$

Then $$\int \frac{du}{2u} = \frac {1}{2} \int \frac {du}{u}$$

u = $$\frac{1}{2} \scriptsize ln(2x + 1)$$

$$\therefore \scriptsize 3 \normalsize \int\frac{1}{x + 5}\scriptsize dx + \normalsize\int \frac{1}{2x + 1}\scriptsize dx$$

= $$\scriptsize3 ln( x + 5) + \normalsize \frac{1}{2} \scriptsize ln(2x + 1) + c$$

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