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

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

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

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

\frac{1}{2} + x - \frac{1}{2} ln 1 + 2 x + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{v - 1}{4 v} d v

= 2 \cdot \int \frac{v - 1}{4 v} d v

Take the constant out:

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

= 2 \cdot \frac{1}{4} \cdot \int \frac{v - 1}{v} d v

Expand

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

= 2 \cdot \frac{1}{4} \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 \cdot \frac{1}{4} (\int 1 d v - \int \frac{1}{v} d v)

\int 1 d v = v

\int \frac{1}{v} d v = ln ∣ v ∣

= 2 \cdot \frac{1}{4} (v - ln ∣ v ∣)

Substitute back

= 2 \cdot \frac{1}{4} (1 + 2 x - ln 1 + 2 x)

Simplify

2 \cdot \frac{1}{4} (1 + 2 x - ln 1 + 2 x) : \frac{1}{2} + x - \frac{1}{2} ln 1 + 2 x

= \frac{1}{2} + x - \frac{1}{2} ln 1 + 2 x

Add a constant to the solution

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

\frac{\partial}{\partial x} (y^{x^{7}})

f (t) = - 16 t^{2} + 32 t + 240, at t = 0

x \to + 2 lim (\frac{x ^{2} - 4}{x - 2})

x^{2}, 4 x

y = 1 + 2 x, y = \frac{3 + x}{3}

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