Three integrals with a quadratic denominator

Text

Evaluate the following integrals, choosing the technique suited to the denominator: a) $\int_0^2\dfrac{dx}{x^2+2x+5}$ b) $\int_0^2\dfrac{dx}{x^2+5x+6}$ c) $\int_0^2\dfrac{dx}{x^2+4x+4}$

Solution

a) Completing the square. $x^2+2x+5=(x+1)^2+4$, so $\int_0^2\frac{dx}{(x+1)^2+4}=\frac12\left[\arctan\frac{x+1}{2}\right]_0^2=\frac12\left(\arctan\frac32-\arctan\frac12\right)\approx0.2596.$

b) Partial fractions. $x^2+5x+6=(x+2)(x+3)$, with $\dfrac{1}{(x+2)(x+3)}=\dfrac{1}{x+2}-\dfrac{1}{x+3}$: $\int_0^2\!\left(\frac{1}{x+2}-\frac{1}{x+3}\right)dx=\left[\ln\frac{x+2}{x+3}\right]_0^2=\ln\frac45-\ln\frac23=\ln\frac65\approx0.1823.$

c) Perfect square. $x^2+4x+4=(x+2)^2$: $\int_0^2\frac{dx}{(x+2)^2}=\left[-\frac{1}{x+2}\right]_0^2=-\frac14+\frac12=\frac14=0.25.$

$\boxed{a=\tfrac12\left(\arctan\tfrac32-\arctan\tfrac12\right),\quad b=\ln\tfrac65,\quad c=\tfrac14}$