Question

Calcule a integral indefinida, utilize a substituição simples:
$\int\ \frac{4x}{ x^{2} -1} \, dx$

Answer 1

Considere u = x² - 1, então:

$du=2xdx~~~\therefore~~~\boxed{\boxed{4xdx=2du}}$

Portanto:

$\int\dfrac{4x}{x^{2}-1}dx=\int\dfrac{2du}{u}=2\int\dfrac{1}{u}du=2\cdot ln~|u| + C=ln~|u|^{2}+C=ln(u)^{2}+C$

Voltando para a variável x:

$\boxed{\boxed{\int\dfrac{4x}{x^{2}-1}dx=ln(x^{2}-1)^{2}+C}}$

Answer 2

$∫ \frac{4x}{ {x}^{2} - 1} dx = 2∫ \frac{2xdx}{ {x}^{2} - 1} \\ u = {x}^{2} - 1 \\ du = 2xdx$

$∫ \frac{2x}{ {x}^{2} - 1} = ln | {x}^{2} - 1 | + c$

$∫ \frac{4x}{ {x}^{2} - 1} = 2ln | {x}^{2} - 1| + c$