How do you integrate $int -x/((x+1)-sqrt(x+1))dx$ ?

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Answer 1 · 2017-02-20T15:34:10

$-x-2sqrt(x+1)+C$

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.

$int -x/((x+1)-sqrt(x+1))dx=-int x/(x-sqrt(x+1)+1)dx$

Let $u=sqrt(x+1)$

$(du)/dx=1/(2sqrt(x+1))$

$dx=2sqrt(x+1) xxdu=2u$ $du$

$x=u^2-1$

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

$-int(u^2-1)/(u^2-1-u+1)2u$ $du=-2int(u(u+1)(u-1))/(u(u-1))$ $du=$

$-2intu+1$ $du=-2(1/2u^2+u)+C=-u^2-2u+C=$

$-(x+1)-2(sqrt(x+1))+C=-x-1-2sqrt(x+1)+C$

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

$-x-2sqrt(x+1)+C$

How do you integrate int -x/((x+1)-sqrt(x+1))dxint -x/((x+1)-sqrt(x+1))dx?