Evaluate the improper integral from zero to infinity.

Question

∫(0 to ∞)1/(x²+3x+2)dx

So I rewrote this as lim(as t approaches infinity) of ∫(0 to t)1/(x^2+3x+2)dx

And then I used integration by parts and factored the denominator so I would get A/(x+2) + B/(x+1)

I got B=1 and A=01, so the equation I ended up integrating -1/(x+1)+1/(x+2).

I got -ln|x+1| + ln|x+2| and by using the fundamental theorem of calculus, I would end up with infinity, therefore making the equation divergent, right?

Answer 1

The partial fraction decomposition of the integrand is 1/(x+1) - 1/(x+2).

Integrating, we get ln l x+1 l - ln l x+2 l = ln l (x+1) / (x+2) l = ln l (1 - 1/(x+2) l (using long division).

Evaluating from 0 to t, we get ln l 1 - 1/(t+2) l - ln (1/2)

Taking the limit as t→∞, we have ln (1 - 0) - ln (1/2) = 0 - [ln(1) - ln(2)] = ln2.

So, the improper integral converges to ln2.