Answer to Question #186257 in Calculus for Phyroe

Question

Improper Integral (Integrals with Infinite Limits)

∫dx/(x^2-2x-3) from 0 to 4

Answer 1

$∫_0^4\dfrac{dx}{(x^2-2x-3)}$

$I =∫_0^4\dfrac{dx}{(x^2-2x-3)}$

We can factorise the integrand and decompose into partial fraction as follows:

$\dfrac{1}{(x^2-2x-3)}\equiv\dfrac{1}{(x-3)(x+1)}$

Thus,

$I =∫_0^4\dfrac{dx}{(x^2-2x-3)}= ∫_0^4\dfrac{\frac{1}{4}}{(x-3)}+\dfrac{\frac{-1}{4}}{(x+1)}$

$I =\dfrac{1}{4}∫_0^4\dfrac{1}{(x-3)}- \dfrac{1}{(x+1)}$

$I = \dfrac{1}{4} \ [ ln|x-3| - ln |x+1|]_0^4$

$I =\dfrac14 \ {(ln1-ln5)-(ln3-ln1)}$

$I = \dfrac{1}{4 }\ (0-ln5-ln3-0)$

$I= -\dfrac{1}{4 }\ (ln5+ln3)$