Integration involving irrational exponent Question:

If: $$I = \int_0^1\frac{dx}{1+x^{\pi/2}}$$ then:
a) $\ln2 < I < \frac{\pi}{4}$
b) $I < \ln2$
c) $I > \frac{\pi}{4}$
d) None of these

I have no idea how to approach this question. How am I supposed to integrate with the irrational exponent of $x$?
 A: Recall that $3 < \pi < 4$ so that $1.5 < \pi/2 < 2$.
(Or directly recall that $\pi/2 = 1.57\ldots$ if this is at your fingertips!)
Well, one technique now is to round down to the integer below, and round up to the integer above.
That is, consider your definite integral where $\pi/2$ is replaced with $1$ and then with $2$.
In each case, compute the result exactly.
How do these results compare to -- i.e., bound! -- the initial integral?
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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Following the @Chip hint, we can perform the integration in terms of the digamma function $\Psi$ as follows:
\begin{align}
I & \equiv \int_{0}^{1}{\dd x \over 1 + x^{\pi/2}} =
\int_{0}^{1}\sum_{n = 0}^{\infty}\pars{-1}^{n}x^{n\pi/2}\,\dd x =
\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over n\pi/2 + 1} =
{2 \over \pi}\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over n + 2/\pi}
\\[3mm] &=
{2 \over \pi}\sum_{n = 0}^{\infty}
\pars{{1 \over 2n + 2/\pi} - {1 \over 2n + 1 + 2/\pi}} =
{2 \over \pi}\sum_{n = 0}^{\infty}
{1 \over \pars{2n + 2/\pi}\pars{2n + 1 + 2/\pi}}
\\[3mm] & =
{1 \over 2\pi}\sum_{n = 0}^{\infty}
{1 \over \pars{n + 1/2 + 1/\pi}\pars{n + 1/\pi}} =
{1 \over 2\pi}\,{\Psi\pars{1/2 + 1/\pi} - \Psi\pars{1/\pi} \over \pars{1/2 + 1/\pi} - 1/\pi}
\\[3mm] & =
\color{#f00}{{1 \over \pi}\,
\bracks{\Psi\pars{\half + {1 \over \pi}} - \Psi\pars{1 \over \pi}}} \approx
0.7533\quad\imp\quad0.6931 = \ln\pars{2} < I < {\pi \over 4} \approx 0.7854
\end{align}
We were not able to get the result by direct manipulation of the digamma involved final result.
A: For $0 < x < 1$, we have $x^2 < x^{\pi/2} <x$. Use these two bound and you can integrate exactly. 
