How do you integrate $\int {(\frac{2}{x + 1})}^{2} d x$ $int( 2 / (x + 1) )^2dx$ ?

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How do you integrate $\int {(\frac{2}{x + 1})}^{2} d x$ $int( 2 / (x + 1) )^2dx$ ?

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Answer 1 · 2016-12-21T21:16:45

$- \frac{4}{x + 1} + C$ $-4/(x + 1) + C$

Explanation:

Let $u = \frac{2}{x + 1}$ $u = 2/(x + 1)$ and $d u = - \frac{2}{{(x + 1)}^{2}} d x$ $du = -2/(x + 1)^2dx$ and $d x = - \frac{{(x + 1)}^{2}}{2} d u$ $dx = -(x + 1)^2/2du$

$\Rightarrow \int {(\frac{2}{x + 1})}^{2} \cdot - \frac{{(x + 1)}^{2}}{2} d u$ $=> int(2/(x + 1))^2 * -(x + 1)^2/2 du$

$\Rightarrow \int (\frac{4}{{(x + 1)}^{2}} \cdot - \frac{{(x + 1)}^{2}}{2}) d u$ $=> int(4/(x + 1)^2 * -(x + 1)^2/2) du$

$\Rightarrow \int (- 2) d u$ $=> int(-2)du$

Use the rule $\int (x^{n}) d x = \frac{x^{n + 1}}{n + 1} + C$ $int(x^n)dx = x^(n + 1)/(n + 1) + C$

$\Rightarrow - 2 u$ $=> -2u$

$\Rightarrow - 2 (\frac{2}{x + 1}) + C$ $=> -2(2/(x + 1)) + C$

$\Rightarrow - \frac{4}{x + 1} + C$ $=> -4/(x + 1) + C$

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How do you integrate $\int {(\frac{2}{x + 1})}^{2} d x$ $int( 2 / (x + 1) )^2dx$ ?

1 Answer

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