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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