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

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

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

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

1 + 2 x - ln 1 + 2 x + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{2 ( v - 1 )}{v} d v

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

Take the constant out:

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

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

Expand

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

= \frac{1}{2} \cdot 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

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

\int 1 d v = v

\int \frac{1}{v} d v = ln ∣ v ∣

= \frac{1}{2} \cdot 2 (v - ln ∣ v ∣)

Substitute back

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

Simplify

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

= 1 + 2 x - ln 1 + 2 x

Add a constant to the solution

= 1 + 2 x - ln 1 + 2 x + C

x \to 0 lim (\frac{( 2 + x ) ^{2} - 4}{h ( x )})

\frac{\partial}{\partial x} (2 - x^{2} + x y - 2 y^{2})

\frac{\partial}{\partial h ( w )} (\frac{w}{h ( w ) ^{2}})

2 e^{5 x} cos (6 x)

\frac{1}{2} (t - 2)^{2}

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