Hi all,
IS this correct?
thanks
John
[math}\int \frac{x.dx}{x^2+4x+13}[/tex]
You've only stated the question? Partial fractions is a good method here but first some algebra:
$\displaystyle \int \dfrac{10}{(2x-6)(x+5)} = \int \dfrac{10}{2(x-3)(x+5)} = 5 \int \dfrac{1}{(x-3)(x+5)}$
$\displaystyle \dfrac{1}{(x-3)(x+5)} = \dfrac{A}{x-3} + \dfrac{B}{x+5} \Leftrightarrow 1 = A(x+5) + B(x-3)$
Can you continue?
edit: That other integral is not an easy one to work out. Are you familiar with trig integrals?
I get B = -1/8
Sub in x=-5: $\displaystyle 1 = B(-5-3) = -8B \Leftrightarrow B = -\dfrac{1}{8}$
Sub in x=3: $\displaystyle 1 = 8A \Leftrightarrow A = \dfrac{1}{8}$
Hence $\displaystyle 5 \int \dfrac{1}{(x-3)(x+5)} = 5 \left(\int \dfrac{1}{8(x-3)} - \int \dfrac{1}{8(x+5)}\right) = \dfrac{5}{8} \left(\int \dfrac{1}{x-3} - \int \dfrac{1}{x+5}\right)$
$\displaystyle =\dfrac{5}{8} \left( \ln(x-3) - \ln(x+5) + C_1\right) = \dfrac{5}{8} \ln \left(\dfrac{x-3}{x+5}\right) + C \text{ where } C = \dfrac{5}{8}C_1$
Perhaps it would be easier to make a new thread since there seem to be two questions on the go here! Alternatively I can delete my posts regarding the original integral