What is the integral of 1/(x^{1/2)+1} ?

The integral of 1/(x^{1/2)+1} is 2(x^{1/2}+1-ln|x^{1/2}+1|)+C

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integral of 1/(x^{1/2)+1}

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

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

Solution steps

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

Apply u-substitution

: \int \frac{2 u}{u + 1} d u

= \int \frac{2 u}{u + 1} 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 + 1} d u

Apply u-substitution

: \int \frac{v - 1}{v} d v

= 2 \cdot \int \frac{v - 1}{v} d v

Expand

\frac{v - 1}{v} : 1 - \frac{1}{v}

= 2 \cdot \int 1 - \frac{1}{v} d v

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 1 d v - \int \frac{1}{v} d v)

\int 1 d v = v

\int \frac{1}{v} d v = ln ∣ v ∣

= 2 (v - ln ∣ v ∣)

Substitute back

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (x y (3 - x - y))

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What is the integral of 1/(x^{1/2)+1} ?
The integral of 1/(x^{1/2)+1} is 2(x^{1/2}+1-ln|x^{1/2}+1|)+C