How do you integrate $int( 2 / (x + 1) )^2dx$ ?

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Answer 1 · 2016-12-21T21:16:45

$-4/(x + 1) + C$

Let $u = 2/(x + 1)$ and $du = -2/(x + 1)^2dx$ and $dx = -(x + 1)^2/2du$

$=> int(2/(x + 1))^2 * -(x + 1)^2/2 du$

$=> int(4/(x + 1)^2 * -(x + 1)^2/2) du$

$=> int(-2)du$

Use the rule $int(x^n)dx = x^(n + 1)/(n + 1) + C$

$=> -2u$

$=> -2(2/(x + 1)) + C$

$=> -4/(x + 1) + C$

Hopefully this helps!

How do you integrate int( 2 / (x + 1) )^2dx int( 2 / (x + 1) )^2dx?