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 $a\in\mathbb{Z}$

  • 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

$$\frac{u^n}{u+a} = u^{n-1} - au^{n-2} + a^2u^{n-3} - \cdots + a^{n-2}u - a^{n-1} -\frac{a^n}{u+a} $$ 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}


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