1
$\begingroup$

Background:

I got stuck in the question below because of a silly error.


Evaluate the integral using the substitution $u=\tan x$ $$\int^{\frac{1}{4}\pi}_0{(\tan^{n+2}x+\tan^{n}x)}\,dx$$


I realized that $du = \sec^2x\,dx$ and hence $du=\mathbf{(u+1)}\,dx$. The error being that $\sec^2x = u^2 +1$, NOT $u+1$.

If we proceed and use the erroneous expression $du = (u+1)\,dx$, we get:

$$\int_0^1{\frac{u^n(u^2+1)}{u+1}}\,du$$ Which is the question that I'm asking: how do you evaluate this integral?

And more generally, how can you evaluate an integral of the form: $$\int{\frac{u^n}{u+a}\,du}$$ where $a\in\mathbb{Z}$

$\endgroup$
  • 3
    $\begingroup$ For the general case, let $u \to u - a$, giving a polynomial of degree $n$ in the numerator and a polynomial of degree $1$ in denominator. $\endgroup$ – Mattos May 14 '17 at 12:01
2
$\begingroup$

$$\frac{u^n}{u+a} = u^{n-1} - au^{n-2} + a^2u^{n-3} - \cdots + a^{n-2}u - a^{n-1} -\frac{a^n}{u+a} $$ So \begin{align} \int \frac{u^n}{u+a} du &= \int \left(u^{n-1} - au^{n-2} + a^2u^{n-3} - \cdots + a^{n-2}u - a^{n-1} -\frac{a^n}{u+a}\right) du\\ &= \frac{u^n}{n} - a \frac{u^{n-1}}{n-1} + a^2 \frac{u^{n-2}}{n-2} - \cdots + a^{n-2} \frac{u^2}{2} - a^{n-1}u - a^n \ln{|u+a|}\\ &= \sum_{m=0}^{n-1} (-1)^ma^m \frac{u^{n-m}}{n-m} - a^n\ln{|u+a|}. \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.