<s>[INST] Using power series representation find an approximation to $\int_{0}^{0.1} \arctan(2x)dx$ Using power series representation find an approximation to 
$$\int_{0}^{0.1} \arctan(2x)dx$$
This is my solution:
$$\frac{d}{dx}\arctan(2x)=\ \frac{2}{1+4x^2} $$
$$\int_{0}^{0.1}\frac{d}{dx}\arctan(2x)=\ \int_{0}^{0.1}\frac{2}{1+4x^2} dx$$
$$\arctan(2x)=\ \int_{0}^{0.1}\frac{2}{1+4x^2} dx$$
$$= \int_{0}^{0.1}2\frac{1}{1-(-4x^2)} \, dx =\int_{0}^{0.1} \sum_{n=0}^\infty 2(-1)^n (4x^2)^n \, dx= \sum_{n=0}^{\infty}\frac{2(-1)^{n}4^{n}0.1^{2n+1}}{2n+1}$$
When I input the formula that I got in the calculator I get a value of 0.1973. But when I input the integral of arctangent I get 0.0099. I guess I missed something and can't find what. Where is my mistake?
 [/INST] We have:
$$ I=\int_{0}^{\frac{1}{10}}\arctan(2x)\,dx = \frac{1}{2}\int_{0}^{\frac{1}{5}}\arctan(x)\,dx \stackrel{IBP}{=} \frac{\arctan\frac{1}{5}}{10}-\frac{\log\left(1+\frac{1}{25}\right)}{4}\tag{1}$$
where, for any $x\in(0,1)$,
$$ \arctan(x)=\sum_{n\geq 0}\frac{(-1)^n x^{2n+1}}{(2n+1)},\qquad \log(1+x^2)=\sum_{n\geq 1}\frac{(-1)^{n+1} x^{2n}}{n}.\tag{2} $$
Due to Leibniz' test,
$$ \left|\arctan\frac{1}{5}-\sum_{n=0}^{1}\frac{(-1)^n}{5^{2n+1}(2n+1)}\right|<\frac{1}{5^6},\qquad \left|\log\left(1+\frac{1}{25}\right)-\sum_{n=1}^{2}\frac{(-1)^{n+1}}{5^{2n}n}\right|<\frac{1}{3\cdot 5^6} $$
hence it follows that:
$$ I \approx \frac{1}{10}\left(\frac{1}{5}-\frac{1}{3\cdot 5^3}\right)-\frac{1}{4}\left(\frac{1}{5^2}-\frac{1}{2\cdot 5^4}\right) = \color{red}{\frac{149}{15000}}=0.00993333\ldots \tag{3}$$
and the approximation error is less than $2\cdot 10^{-5}$.
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