It's rational integral

Claim: $\int \frac{x^4+2x+2}{x^5+x^4}\;\mathrm{d} x=\ln(|x+1|)-\frac{2}{3x^3}+C$ Proof: $\begin{aligned} \int \frac{x^4+2x+2}{x^5+x^4}\;\mathrm{d} x&=\int \frac{x^4+2(x+1)}{x^4(x+1)}\;\mathrm{d} x \\ \\ &=\int \left(\frac{1}{x+1}+\frac{2}{x^4} \right)\;\mathrm{d}x \\ \\ &=\int \frac{1}{x+1}\;\mathrm{d}x+\int \frac{2}{x^4}\;\mathrm{d}x \\ \\ &=\ln(|x+1|)-\frac{2}{3x^3}+C \end{aligned}$ Evaluating the given definite integral, we have, $\begin{aligned} \int^{2}_1 \frac{x^4+2x+2}{x^5+x^4}\;\mathrm{d} x&= \left[\ln(|x+1|)-\frac{2}{3x^3}\right]^{2}_1 \\ \\ &=\left(\ln(3)-\frac{1}{12}\right)-\left(\ln(2)-\frac{2}{3}\right) \\ \\ &=\ln\left(\frac{3}{2}\right)+\frac{7}{12} \end{aligned}$ Therefore, $\;\dfrac{2n+1}{4n}=\dfrac{7}{12}\;$ and $\;\ln\left(\dfrac{n}{n-1}\right)=\ln\left(\dfrac{3}{2}\right)\implies \boxed{n=3}$

$\begin{aligned} I & = \int_1^2 \frac {x^4+2x+2}{x^5+x^4} dx \\ & = \int_1^2 \frac {x^4 + 2(x+1)}{x^4(x+1)} dx \\ & = \int_1^2 \left(\frac 1{x+1} + \frac 2{x^4} \right) dx \\ & = \ln (x+1) - \frac 2{3x^3} \ \bigg|_1^2 \\ & = \ln 3 - \frac 1{12} - \ln 2 + \frac 23 \\ & = \frac 7{12} - \ln \frac 32 \end{aligned}$

Therefore $n=\boxed 3$ .