All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.


Using partial fractions formula found in MF15,

Let \frac{9x^{2}+x-13}{(2x-5)(x^{2}+9)} = \frac{A}{2x-5} + \frac{Bx+C}{x^{2}+9}

9x^{2}+x-13 = A(x^{2}+9)+(Bx+C)(2x-5)

9 = A + 2B \longrightarrow 1

1 = 2C - 5B \longrightarrow 2

-13 = 9A - 5C \longrightarrow 3

Using GC, A = 3, B = 3, C = 8

\int_{0}^{2} \frac{9x^{2}+x-13}{(2x-5)(x^{2}+9)} dx

= \int_{0}^{2} \frac{3}{2x-5} + \frac{3x+8}{x^{2}+9} dx

= \int_{0}^{2} \frac{3}{2} \frac{2}{2x-5} + \frac{3x}{x^{2}+9} + \frac{8}{x^{2}+9}dx

= \int_{0}^{2} \frac{3}{2} \frac{2}{2x-5} + \frac{3}{2} \frac{2x}{x^{2}+9} + 8 \frac{1}{x^{2}+9}dx

= [\frac{3}{2} \mathrm{ln}|2x-5|+ \frac{3}{2} \mathrm{ln}(x^{2}+9) + \frac{8}{3} tan^{-1} \frac{8}{3}]\biggl|_0^2

= \frac{3}{2} \mathrm{ln} 13 + \frac{8}{3} tan^{-1} \frac{2}{3} - (\frac{3}{2} \mathrm{ln}5 + \frac{3}{2} \mathrm{ln}9]

= \frac{3}{2} \mathrm{ln} \frac{13}{45} + \frac{8}{3} tan^{-1} \frac{2}{3}

a = \frac{3}{2}, b = \frac{13}{45}, c = \frac{8}{3}, d = \frac{2}{3}

Personal Comments:
Firstly, this question is a lot of marks! It is quite a standard tutorial question that tests students on all their integration techniques. I think that they combined system of linear equations here too, which is really neat. Students could also have solved the partial fractions using the substitution (cover-up) method. Do use the MF15 for the partial fractions and integration formula!
Finally, the answers are to be left in rational form, which means to be expressed as a fraction of two integers!

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