What is the integral of 1/(x-sqrt(x+2)) ?

The integral of 1/(x-sqrt(x+2)) is 2(1/3 ln|sqrt(x+2)+1|+2/3 ln|sqrt(x+2)-2|)+C

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integral of 1/(x-sqrt(x+2))

\int \frac{1}{x - x + 2} d x

2 (\frac{1}{3} ln x + 2 + 1 + \frac{2}{3} ln x + 2 - 2) + C

Solution steps

\int \frac{1}{x - x + 2} d x

Apply u-substitution

: \int \frac{2 u}{u ^{2} - 2 - u} d u

= \int \frac{2 u}{u ^{2} - 2 - u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{u}{u ^{2} - 2 - u} d u

Take the partial fraction of

\frac{u}{u ^{2} - 2 - u} : \frac{1}{3 ( u + 1 )} + \frac{2}{3 ( u - 2 )}

= 2 \cdot \int \frac{1}{3 ( u + 1 )} + \frac{2}{3 ( u - 2 )} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int \frac{1}{3 ( u + 1 )} d u + \int \frac{2}{3 ( u - 2 )} d u)

\int \frac{1}{3 ( u + 1 )} d u = \frac{1}{3} ln ∣ u + 1 ∣

\int \frac{2}{3 ( u - 2 )} d u = \frac{2}{3} ln ∣ u - 2 ∣

= 2 (\frac{1}{3} ln ∣ u + 1 ∣ + \frac{2}{3} ln ∣ u - 2 ∣)

Substitute back

u = x + 2

= 2 (\frac{1}{3} ln x + 2 + 1 + \frac{2}{3} ln x + 2 - 2)

Add a constant to the solution

= 2 (\frac{1}{3} ln x + 2 + 1 + \frac{2}{3} ln x + 2 - 2) + C

y = x^{2}, y = 4^{\frac{2}{3}}

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What is the integral of 1/(x-sqrt(x+2)) ?
The integral of 1/(x-sqrt(x+2)) is 2(1/3 ln|sqrt(x+2)+1|+2/3 ln|sqrt(x+2)-2|)+C