부분분수 적분

The integrand is a rational function, so it's a candidate for partial fraction. We quickly rule out any obvious substitution or integration by parts, so we go ahead with the decomposition.

We start by expressing the integrand, i.e. the fraction $\frac{1}{x+x^2}=\frac{1}{x(1+x)}$, as a linear combination of the simpler fractions $\frac{1}{x}$ and $\frac{1}{x+1}$ (which we already know how to integrate). We will have

$$\begin{aligned} \frac{1}{x+x^2}=\frac{1}{x(1+x)} = \frac{a}{x} + \frac{b}{x+1} =\frac{a(x+1) +bx}{x(1+x)} \end{aligned}$$

The fraction on the left hand side is the same as the fraction on the right hand side if and only if the numerator on the left hand side, which is $1 = 0x + 1$, is equal to the numerator on the right hand side, which is $a(x+1)+bx = (a+b)x +a$. This in turn is the case if and only of $a=1$ (i.e. the constant terms are the same in the two numerators) and $a+b=0$ (i.e. the coefficients of $x$ are the same in the two numerators). So $a=1$ and $b=-1$. Now we can easily evaluate the integral

$$\begin{aligned} \int_1^2 \frac{\mathrm{d}x}{x+x^2} &=\int_1^2 \frac{\mathrm{d}x}{x(x+1)} =\int_1^2 \Big(\frac{1}{x}-\frac{1}{x+1}\Big)\,\mathrm{d}x =\Big[\log x-\log(x+1)\Big]_1^2\\ &=\log 2-\log\frac{3}{2} =\log\frac{4}{3} \end{aligned}$$