How to estimate definite integral using Taylor series part

Question

I've worked out the Taylor series for $y(x)=\ln (1+2x)$ "up to and including terms of 0(x5)": $$2x - 2x^2 + {8x^3\over 3} - 4x^4 + {32x^5\over 5}$$ Now I have to use the series to estimate the value of the definite integral $$\int_0^\frac 14 \frac 1x\ln (1+2x)dx$$ Every method I've found has involved a series with a formula involving only 'x' and 'n' as variables, while the $\ln (1+2x)$ series is represented by $$\sum_{n=0}^\infty {f^n (0)x^n\over n!}$$ I am unsure both how to incorporate the new $\frac 1x$ into the Taylor series I'm using, and how to estimate the definite integral with it.

Answer 1

$$\ln (1+2x)\approx 2x - 2x^2 + {8x^3\over 3} - 4x^4 + {32x^5\over 5}$$ Multiplying by $\frac{1}{x}$ $$\frac{1}{x}\ln (1+2x)\approx 2- 2x + {8x^2\over 3} - 4x^3 + {32x^4\over 5}$$$$$$

$$\int_0^{\frac{1}{4}} \frac{1}{x} \ln (1+2x) \approx \int_0^{\frac{1}{4}} ( 2-2x+\frac{8x^2}{3}-4x^3+\frac{32x^4}{5})$$$$\int_0^{\frac{1}{4}} \frac{1}{x} \ln (1+2x) \approx \Biggr[2x-x^2+\frac{8x^3}{9}-x^4+\frac{32x^5}{25}\Biggr]_0^{\frac{1}{4}}$$$$\int_0^{\frac{1}{4}} \frac{1}{x} \ln (1+2x) \approx 0.448733$$

It is a good estimate,