By expanding the integrand in a power series, calculate

By expanding the integrand in a power series, calculate $\displaystyle\int_0^{1/2}\frac{\arctan x}x\,dx$ with accuracy $10^{-2}$.

Explanation of Method

Dividing the $\arctan x$ series by $x$ and integrating over $[0,\frac12]$ gives an alternating series with an extra factor of $2^{-(2n+1)}$, which converges rapidly enough that only a few terms are needed.

Divide the $\arctan x$ series by $x$.
Integrate term by term over $[0,\frac12]$.
Sum until the error bound meets the tolerance.

Step 1 · Divide by $x$

$$ \frac{\arctan x}x = \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{2n+1}. $$

Step 2 · Integrate over $[0,\frac12]$

$$ \int_0^{1/2}\frac{\arctan x}x\,dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2\cdot2^{2n+1}}. $$

Step 3 · Sum the terms

$$ \begin{aligned} n=0:&\ 0.500000 \\ n=1:&\ -0.013889 \\ n=2:&\ 0.001250 \end{aligned} $$ Sum through $n=2$: $0.5-0.013889+0.001250=0.487361$, with next-term error $\approx0.000159<10^{-2}$.

Summary Table

Property	Value
Series	$\sum(-1)^n/[(2n+1)^2\cdot2^{2n+1}]$
Result	$0.49$
Terms needed	$3$

Additional Notes

The extra factor $2^{-(2n+1)}$ from the small upper limit $1/2$ makes this converge much faster than the $x=1$ dilogarithm-type integrals of Problems 264–265, needing only three terms for two-decimal accuracy.
Structurally this is a scaled analogue of Problem 265's $\int_1^x\arctan(t)/t\,dt$, but with the integration starting at $0$ rather than $1$, avoiding any boundary subtlety with Catalan's constant.
Even using only terms $n=0,1$ (giving $0.4861$) would already meet the $10^{-2}$ tolerance, showing how conservative the full three-term computation is.

Result

$$ \int_0^{1/2}\frac{\arctan x}x\,dx \approx 0.49, \quad \text{error} < 10^{-2}. $$