calculus - How to prove $int_0^1 frac{1+x^{30}}{1+x^{60}} dx = 1 + frac{c}{31}$, where $0 < c < 1$

This is an exercise from Apostol (p.285) that I'm having trouble with (in fact, I'm having trouble with the whole section):

Prove that $\displaystyle{\int_0^1 \frac{1+x^{30}}{1+x^{60}} = 1 + \frac{c}{31}}, \qquad \text{where } 0 < c < 1.$

This comes from the section of Exercises following Taylor expansions, and Taylor's formula with error term. It seems like the approach should involve getting this as the error term of the Taylor expansion of a function that we know something about? I'm having trouble making much more progress than that.

Updated Progress: From the definition of the error term of the Taylor expansion we have:

$$E_n (x) = \frac{1}{n!} \int_a^x (x-t)^n f^{(n+1)} (t) dt$$

Alternatively, we may express this in the Lagrange form of the error term (derived from the weighted mean value theorem of integrals):

$$E_n (x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x-a)^{n+1} \qquad \text{where } a < c < x.$$

Now, let $a = 0$, $x = 1$, $n = 30$, $f^{(n+1)}(t) = \dfrac{1+t^{30}}{(1+t^{60})(1-t)^{30}}$, and set the two alternative expressions of $E_n(x)$ equal to each other:

$$\begin{align*}
\implies & \frac{1}{n!} \int_a^x (x-t)^n f^{(n+1)}(t) dt & = & \dfrac{f^{(n+1)}(c)}{(n+1)!} (x-a)^{n+1}\\
\implies & \frac{1}{30!} \int_0^1 (1-t)^{30} \dfrac{1+t^{30}}{(1+t^{60})(1-t)^{30}} & = & \dfrac{1}{31!} \dfrac{1+c^{30}}{(1+c^{60})(1-c)^{30}}\\
\implies & \int_0^1 \dfrac{1+t^{30}}{1+t^{60}} & = & \frac{1}{31} \frac{1+c^{30}}{(1+c^{60})(1-c)^{30}}.
\end{align*}
$$

I cannot seem to get from here to $=1 + \dfrac{c}{31}$.

If someone can show me pretty explicitly, step-by-step how to tackle this problem, I'd appreciate it. Dealing with the error term on Taylor expansions is giving me quite a bit of trouble, so hopefully seeing a full solution will help clear things up.

Answer

The approach of anon is better, but we can use the power series expansion of $\frac{1}{1+x^{60}}$ to conclude that for $-1\le x\le 1$,
$$\frac{1+x^{30}}{1+x^{60}}=(1+x^{30})(1-x^{60}+x^{120}-x^{180}+\cdots).$$
Multiply out, and integrate term by term. We get that our integral $I$ is given by
$$I=1+\frac{1}{31}-\frac{1}{61}-\frac{1}{91}+\frac{1}{121}+\cdots.$$
Group terms by twos, either starting at the beginning, or starting after the first term. From the two groupings, we can see that
$$1We can use the same idea to get a string of increasingly more precise inequalities, such as
$$1+\frac{1}{31}-\frac{1}{61}-\frac{1}{91}