Integration by Partial Fraction

Topic Content:

Expressions such as \( \int \frac{7x \: + \:8}{2x^2 \: + \: 11x \: + \: 5}\scriptsize \:dx \)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 \( \scriptsize 7 \left (-\frac{1}{2} \right) \: + \: 8 = A \left [2 \left(-\frac{1}{2} \right) \: + \: 1 \right] \: + \: B\left(-\frac{1}{2} \: + \: 5 \right)\)

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

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

= \( \scriptsize 4 \frac{1}{2} = B 4 \frac{1}{2} \)

\( \frac{9}{2} = \frac{9}{2}\scriptsize B\)

\( \scriptsize B = \normalsize \frac{9}{2} \; \times \frac{2}{9}\)

\(\therefore \scriptsize B = 1 \)

∴ \(\frac{7x \: + \: 8}{2x^2 \: + \: 11x \: + \: 5} = \frac{3}{x \: + \: 5} \: + \: \frac{1}{2x \: + \: 1} \)

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

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