Evaluate: $\int \frac{t^5-t^3}{t^{10}-1}\, dt. $ None of my attempts (such as letting $x=t^5$ or splitting up the integrand into two parts etc.) succeeded. Basically I couldn't find a useful thing on manipulating the given integrand. Please give me some hints.. Thanks in advance. 
 A: With $u=t^2$ this is $\tfrac12\int\frac{u(u-1)\,\mathrm{d}u}{u^5-1}=\tfrac12\int\frac{u\,\mathrm{d}u}{u^4+u^3+u^2+u+1}$. You can factorize the quartic denominator into two real quadratics, using fifth roots of unity. Then you can proceed with partial fractions. (You might not like the result if you don't remove the explicit role of complex numbers early. See @AlexeyBurdin's comment.)
A: $$
\int \frac{t^5-t^3}{t^{10}-1}\, dt = \int \frac{t^4-t^2}{t^{10}-1} \, \big( t\, dt\big)
$$
Why would I do what you see above? Note that $t$ is a constant multiple of the derivative of $t^2,$ and doing what appears above leaves you with a function of $t^2$ in the fraction. In other words, this substitution is indicated:
\begin{align}
& u = t^2 \\
& du = 2t\, dt \\
& du/2 = t\,dt
\end{align}
So the integral becomes
$$
\int \frac{u^2 -u}{u^5 -1} \cdot \frac{du} 2
$$
Now to use a partial fraction decomopsition you need to factor $u^5-1.$ Since $u^5-1=0$ when $u=1,$ you have $u-1$ as a factor, thus:
$$
u^5-1= (u-1)(u^4+u^3+u^2+1)
$$
However, this leaves the problem of how to factor that 4th-degree polynomial. There's a formula for factoring 4th-degree polynomials, but it's cumbersome, and what most mathematicians know about it is where to look it up in a book. But in this case, the way to do this is to look at the other solutions of $u^5-1=0$ besides $u=1.$
We need $u^5 = 1.$ Necessarily this means $|u|=1,$ so the values of $u$ that we seek must be on the circle centered at $0$ with radius $1.$ That means they are of the form $\cos\alpha+i\sin\alpha.$ Now recall that
$$
\big( \cos\alpha+i\sin\alpha\big)^5 = \cos(5\alpha) + i \sin(5\alpha).
$$
This is equal to $1$ only if $\cos(5\alpha)=1$ and $\sin(5\alpha)=0.$
So we need $\alpha$ to be one-fifth of an integer multiple of $360^\circ.$ Thus $\alpha\in\{0^\circ,\pm72^\circ, \pm144^\circ\}$ and we've already found the case $\alpha=0^\circ$ with $u=1.$ Looking at $\pm72^\circ,$ we have
\begin{align}
& \Big( x - \big(\cos72^\circ+i\sin72^\circ\big) \Big) \Big( x - \big( \cos72^\circ - i\sin72^\circ \big) \Big) \\
& \text{complex-conjugate pairs, so the imaginary parts cancel out} \\[10pt]
= {} & x^2 - 2x\cos72^\circ + 1.
\end{align}
And similarly with $144^\circ.$ So we get:
$$
u^5-1 = (u-1) (x^2 - 2x\cos72^\circ + 1) (x^2 - 2x\cos144^\circ + 1)
$$
After that, use partial fractions.
A: And I'll finish the thing. $I=\int\frac{t^5-t^3}{t^{10}-1}\,\mathrm{d}t$
Let $u=t^2,du=2t\,dt$, by J.G.'s answer
$$I=\frac12\int\frac{u\,\mathrm{d}u}{u^4+u^3+u^2+u+1}$$
Using $$u^4+u^3+u^2+u+1=\left(u^2+\frac{1+\sqrt{5}}{2}u+1\right)
\left(u^2+\frac{1-\sqrt{5}}{2}u+1\right)$$ both parenthesis have no real roots, so we will have to boil that down to $\int\frac{ax+b}{x^2+1}\mathrm{d}x$ for some $x$ after partial decomposition. So complete the squares:
$$\left(u^2+\frac{1+\sqrt{5}}{2}u+1\right)
\left(u^2+\frac{1-\sqrt{5}}{2}u+1\right)=\\
\left(\left(\frac{4u+1+\sqrt{5}}{4}\right)^2+\frac{5-\sqrt{5}}{8}\right)
\left(\left(\frac{-4u-1+\sqrt{5}}{4}\right)^2+\frac{5+\sqrt{5}}{8}\right)$$
As we see $4u+1$ shows up in both cases, let $4u+1=x$, $u=\frac{x-1}{4}$, $\mathrm{d}u=\frac{1}{4}\mathrm{d}x$ and
$$I=8\int\frac{x-1}{x^4 + 10 x^2 + 40 x + 205}\mathrm{d}x$$
$$=8\int\frac{x-1}{\left((x - \sqrt{5})^2 + 2\sqrt{5} + 10\right)
\left((x + \sqrt{5})^2 - 2\sqrt{5} + 10\right)}\mathrm{d}x$$
and the partial fraction decomposition is (details are here and here)$$
\frac{x-1}{x^4 + 10 x^2 + 40 x + 205}=
\frac{1}{4\sqrt{5}}\left(\frac{1}{(x - \sqrt{5})^2 + 2\sqrt{5} + 10}-\frac{1}{(x + \sqrt{5})^2 - 2\sqrt{5} + 10}\right)$$
$$I=\frac{2\sqrt{5}}{5}\left(
\frac{1}{\sqrt{2\sqrt{5} + 10}}\int\frac{\mathrm{d}\left(\frac{x-\sqrt{5}}{\sqrt{2\sqrt{5} + 10}}\right)}{\left(\frac{x-\sqrt{5}}{\sqrt{2\sqrt{5} + 10}}\right)^2+1}-
\frac{1}{\sqrt{-2\sqrt{5} + 10}}\int\frac{\mathrm{d}\left(\frac{x+\sqrt{5}}{\sqrt{-2\sqrt{5} + 10}}\right)}{\left(\frac{x+\sqrt{5}}{\sqrt{-2\sqrt{5} + 10}}\right)^2+1}
\right)$$
$$=\frac{2\sqrt{5}}{5}\left(
\frac{1}{\sqrt{2\sqrt{5} + 10}}\arctan\left(\frac{x-\sqrt{5}}{\sqrt{2\sqrt{5} + 10}}\right)-
\frac{1}{\sqrt{-2\sqrt{5} + 10}}\arctan\left(\frac{x+\sqrt{5}}{\sqrt{-2\sqrt{5} + 10}}\right)
\right)+C$$
where $x=4t^2+1$.
