Evaluate by polynomial expansion the following integral with an error less than 0.000001 My work till now:
$\int_0^{0.2} \frac{1}{1+x^5} dx=\int_0^{0.2}(1+x^5)^{-1} dx$
$(1+x)^n=1+nx+ \frac{n(n-1)x^2}{2}...$
$\int_0^{0.2} \bigl( 1+(-1)x^5+ \frac{(-1)(-2)x^{10}}{2}- \frac{1(-2)(-3)x^{15}}{3!}+...\bigr)dx$
Now, I could integrate the above series but how can I find the result within that error?
Any tip/help is appreciated. Thanks!

I think that the approximation works well: since
$$1+x^{15}=(1+x^5)(1-x^5+x^{10})$$
it follows that for any $x\in [0,1]$
$$\frac{1}{1+x^5}=1-x^5+x^{10}-\frac{x^{15}}{1+x^5}.$$
Hence
$$\begin{align}
\left|\int_{0}^{0.2}\frac{1}{1+x^5}\,dx-\int_{0}^{0.2}(1-x^5+x^{10})\,dx\right|&\leq\int_{0}^{0.2} \frac{x^{15}}{1+x^5}\,dx\\
&\leq \int_{0}^{0.2} x^{15}\,dx=\frac{(0.2)^{16}}{16}<10^{-12}.
\end{align}$$
Note that
$$\int_{0}^{0.2}\frac{1}{1+x^5}\,dx
\approx 0.199989335194742$$
and
$$\int_{0}^{0.2}(1-x^5+x^{10})\,dx\approx 0.1999893351951515.$$
P.S. Actually for an error less that $0.000001=10^{-6}$ you may consider less terms:
$$\frac{1}{1+x^5}=1-x^5+\frac{x^{10}}{1+x^5}.$$
