Question

Calcule a integral indefinida:
$\displaystyle \int\,[x(x^n+a^n)]^{-1}\,dx$

Answer 1

Vamos fazer uma substituição:

$I=\displaystyle\int [x(x^n+a^n)]^{-1}\,dx\\\\ y=(x^n+a^n)\Longrightarrow dy=nx^{n-1}dx\Longrightarrow dx=\dfrac{dy}{nx^{n-1}}$ $[x(x^n+a^n)]^{-1}dx=[xy]^{-1}\cdot\dfrac{dy}{nx^{n-1}}=\dfrac{dy}{nx^ny}$ $I=\displaystyle\int\dfrac{dy}{n(y-a^n)y}=\dfrac{1}{n}\displaystyle\int\dfrac{dy}{(y-a^n)y}$

Podemos usar o método de frações parciais:

$\dfrac{1}{y(y-a^n)}=\dfrac{A}{y}+\dfrac{B}{y-a^n}=\dfrac{A(y-a^n)+By}{y(y-a^n)}=\dfrac{(A+B)y-Aa^n}{y(y-a^n)}\\\\ \begin{cases}A+B=0\to A=-B\to B=\dfrac{1}{a^n}\\-Aa^n=1\to A=-\dfrac{1}{a^n}\end{cases}$

Agora, substituindo o que foi obtido acima:

$I=\dfrac{1}{n}\left(\displaystyle\int\dfrac{-\frac{1}{a^n}}{y}dy+\displaystyle\int\dfrac{\frac{1}{a^n}}{y-a^n}dy\right)\\\\ I=\dfrac{1}{na^n}\left(\displaystyle\int\dfrac{-1}{y}dy+\displaystyle\int\dfrac{1}{y-a^n}dy\right)\\\\ I=\dfrac{1}{na^n}(-\ln(y)+\ln(y-a^n)\\\\ I=\dfrac{1}{na^n}(\ln(x^n)-\ln(x^n+a^n))\\\\ \boxed{I=\dfrac{\ln(x)}{a^n}-\dfrac{\ln(x^n+a^n)}{na^n}+C}$