Question

Integration by Parts Fractions

1.) ∫(8t^3 + 13)dt/(t+2)(4t^2+1)

2.) ∫(5x^2 -3x+ 18)dx/x(9-x^2) from 1 to 2

Answer 1

$\displaystyle (1)\\ \frac{8t^3 + 13}{(t+2)(4t^2+1)}= 2 - \frac{3}{2 + t} - {4t - 6}{1 + 4 t^2}\\ \begin{aligned} \int \frac{8t^3 + 13}{(t+2)(4t^2+1)}\,\mathrm{d}t &= \int 2 - \frac{3}{2 + t} - \frac{4t - 6}{1 + 4 t^2}\,\mathrm{d}t \\& = 2t - 3\ln(2 + t) \\& - \frac{\ln(1 + 4t^2)}{2} + 3\arctan(2t) + C \end{aligned}\\ (2)\\ \frac{5x^2 -3x+ 18}{x(9-x^2)} = \frac{2}{x} - \frac{4}{x + 3} - \frac{3}{x - 3}\\ \begin{aligned} \int\frac{5x^2 -3x+ 18}{x(9-x^2)} \,\mathrm{d}x &= \int\frac{2}{x}\mathrm{d}x - \int\frac{4}{x + 3} + \int \frac{3}{3 - x}\mathrm{d}x \\&= 2\ln{x} - 4\ln(x + 3) - 3\ln(3 - x) + C\\ \int_1^2\frac{5x^2 -3x+ 18}{x(9-x^2)} \,\mathrm{d}x &= 2\ln(2) - 4\ln(5) + 4\ln(4) + 3\ln(2) \\& = 13\ln(2) - 4\ln(5) \end{aligned}$