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

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

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Answer 1 · 2017-02-20T15:34:10

$- x - 2 \sqrt{x + 1} + C$ $-x-2sqrt(x+1)+C$

Explanation:

In order to integrate this, we need to use a method called u-substitution, but a good idea is to remove all constants out of the integral first.

$\int - \frac{x}{(x + 1) - \sqrt{x + 1}} d x = - \int \frac{x}{x - \sqrt{x + 1} + 1} d x$ $int -x/((x+1)-sqrt(x+1))dx=-int x/(x-sqrt(x+1)+1)dx$

Let $u = \sqrt{x + 1}$ $u=sqrt(x+1)$

$\frac{d u}{d x} = \frac{1}{2 \sqrt{x + 1}}$ $(du)/dx=1/(2sqrt(x+1))$

$d x = 2 \sqrt{x + 1} \times d u = 2 u$ $dx=2sqrt(x+1) xxdu=2u$ $d u$ $du$

$x = u^{2} - 1$ $x=u^2-1$

The integral can now be rewritten in terms of $u$ $u$

$- \int \frac{u^{2} - 1}{u^{2} - 1 - u + 1} 2 u$ $-int(u^2-1)/(u^2-1-u+1)2u$ $d u = - 2 \int \frac{u (u + 1) (u - 1)}{u (u - 1)}$ $du=-2int(u(u+1)(u-1))/(u(u-1))$ $d u =$ $du=$

$- 2 \int u + 1$ $-2intu+1$ $d u = - 2 (\frac{1}{2} u^{2} + u) + C = - u^{2} - 2 u + C =$ $du=-2(1/2u^2+u)+C=-u^2-2u+C=$

$- (x + 1) - 2 (\sqrt{x + 1}) + C = - x - 1 - 2 \sqrt{x + 1} + C$ $-(x+1)-2(sqrt(x+1))+C=-x-1-2sqrt(x+1)+C$

Since $1$ $1$ is a constant, we can add it to $C$ $C$ , leaving the answer as:

$- x - 2 \sqrt{x + 1} + C$ $-x-2sqrt(x+1)+C$

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

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Explanation:

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