Question.
How to evaluate
$$
I = \int\frac{x^{5}}{x-1}\,dx?
$$
First of all, let me tell that this integral is very easy, but still I am asking about this integral because I want an alternative solution to this integral.
Basically to integrate this above expression we have to substitute $(x-1)=u$ so $dx=du$ and $x=(u+1)$. Therefore, we can write this above integral as:
$$
\begin{align}
I &= \int\frac{x^{5}}{x-1}\,dx \\
&= \int\frac{(u+1)^{5}}{u}\,du \\
&= \int\frac{u^{5}+5u^{4}+10u^{3}+10u^{2}+5u+1}{u}\,du \\
&= \int\Bigl(u^{4}+5u^{3}+10u^{2}+10u+5+\frac{1}{u}\Bigr)\,du.
\end{align}
$$
Now after this step we can easily integrate this above expression.
My doubt.
Is there any other way to integrate this above expression without writing the binomial expansion of $(1+u)^{5}$? For e.g. here in this integral since the highest power of $x$ is $5$ so it is easy to integrate after writing the binomial expansion. But if this question was
Integrate
$$
I = \int\frac{x^{15}}{x-1}\,dx,
$$
then it would have been very tough to integrate by writing the binomial expansion of $(1+u)^{15}$ after substituting $(x-1)=u$. So, I want to know whether there is any shortcut tricks to evaluate these types of integrals.
Best Answer
Expanding on the hint by @AnuragA in the comments, a rational function can always be rewritten as the sum of a polynomial (quotient) and a proper fraction (from the remainder) using the polynomial division algorithm.
As a consequence, we can rewrite $$ \frac{A(x)}{B(x)} = Q(x) + \frac{R(x)}{B(x)}. $$
Why is this "better" for integration? Obviously, $Q(x)$ is a polynomial, so that's nice. But when we look at degrees, the remaining rational term is a proper fraction, so it will be easier to deal with in general. If the denominator factors, then partial fraction decomposition can be applied to it.
Moreover, since $\deg R(x) < \deg B(x)$, the remainder fraction $\smash{\frac{A(x)}{B(x)}}$ term tends to $0$ as $x \to \pm \infty$, so that asymptotically, the original rational function behaves like the polynomial $Q(x)$. This describes all the types of asymptotes (horizontal, "slant" i.e. linear, and otherwise).
In your example (and its generalization), since $B(x) = x - 1$ is linear (degree $1$), the remainder has to be constant (degree $0$), which means that the antiderivative will just consist of polynomial terms together with a single logarithm term.