Express $\frac{3x^2-5x-3}{x^3-3x^2}$ in partial fractions. Hence, evaluate \(\displaystyle\int_{4}

Question

Express $\frac{3x^2-5x-3}{x^3-3x^2}$ in partial fractions. Hence, evaluate $\displaystyle\int_{4}^{6} \frac{3x^2-5x-3}{x^3-3x^2} \, \mathrm{d}x$.

Accepted Answer

$\frac{3x^2-5x-3}{x^3-3x^2}=\frac2x+\frac1{x^2}+\frac1{x-3}$, and hence $\int_4^6\frac{3x^2-5x-3}{x^3-3x^2}\,\mathrm dx=2\ln6+\ln3-4\ln2+\frac1{12}$

Techniques of Integration - Integration Using Partial Fractions