How can I solve the improper integral $\int_{1}^\infty {dx \over {(x+1)(x+2)}}$

Question

$$\int_{1}^\infty {dx \over {(x+1)(x+2)}}$$

I have the indefinite integral solved for:

$$\ln(x+1)-\ln(x+2) + C$$

But I don't know how to finish with $[1, \infty]$.

Answer 1

Note that $$\ln(x+1)-\ln(x+2)=\ln\frac{x+1}{x+2}=\ln\frac{1+\frac 1x}{1+\frac 2x}.$$

Answer 2

Hint: split the fraction in two, then use the improper integral

Answer 3

Use

$$\ln(x+1)-\ln(x+2)=\ln\left(\frac{x+1}{x+2}\right)$$

Answer 4

$$\ln (\frac{x+1}{x+2})=\ln(\frac{1+1/x}{1+2/x})$$

Answer 5

To find the "value" at $+ \infty$, notice that

$$\ln(x+1) - \ln(x+2) = \ln \left( \frac{x+1}{x+2} \right)$$

From here the limit is easy to calculate

Answer 6

$\ln(x+1)-\ln(x+2) + C = \ln(\frac{x+1}{x+2}) + C$ evaluated at 1 and $\infty$ gives $\ln(\frac{\infty}{\infty}) - \ln\frac{2}{3}=\ln\frac{3}{2}$

Answer 7

$$\int_{1}^{M}\frac{dx}{(x+1)(x+2)}=\int_{1}^{M}\frac{dx}{x+1}-\int_{1}^{M}\frac{dx}{x+2}=\int_{1}^{M}\frac{dx}{x+1}-\int_{2}^{M+1}\frac{dx}{x+1}$$ hence: $$\int_{1}^{M}\frac{dx}{(x+1)(x+2)}=\int_{1}^{2}\frac{dx}{x+1}-\int_{M}^{M+1}\frac{dx}{x+1}=\log\frac{3}{2}+O\left(\frac{1}{M}\right).$$