$\int\limits_1^2\frac{1}{x(x+1)}dx,x>0$ equals
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\log(\frac{4}{3})$
Integral: $\int_1^2 \frac{1}{x(x+1)} \, dx$
Use partial fractions: $\frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1}$
$1 = A(x+1) + Bx \Rightarrow 1 = A x + A + B x = (A+B)x + A$
Compare coefficients: $A + B = 0$, $A = 1 \Rightarrow B = -1$
Integral: $\int_1^2 \left(\frac{1}{x} - \frac{1}{x+1}\right) dx = \int_1^2 \frac{1}{x} dx - \int_1^2 \frac{1}{x+1} dx$
$= [\ln x]_1^2 - [\ln(x+1)]_1^2 = (\ln 2 - \ln 1) - (\ln 3 - \ln 2) = \ln 2 - (\ln 3 - \ln 2) = 2 \ln 2 - \ln 3 = \ln \frac{4}{3}$
Answer: $\ln \frac{4}{3}$