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SS3: MATHEMATICS - 2ND TERM

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  1. Matrices I | Week 1
    6 Topics
  2. Matrices II | Week 2
    1 Topic
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    1 Quiz
  3. Commercial Arithmetic | Week 3
    7 Topics
    |
    1 Quiz
  4. Coordinate Geometry | Week 4
    8 Topics
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    1 Quiz
  5. Differentiation of Algebraic Expressions | Week 5 & 6
    7 Topics
  6. Application of Differentiation | Week 7
    4 Topics
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    1 Quiz
  7. Integration | Week 8
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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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