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

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

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

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

2 ln 1 + x + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 2 ln ∣ v ∣

Substitute back

= 2 ln 1 + x

Add a constant to the solution

= 2 ln 1 + x + C

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

\frac{d}{d x} (x^{3} - \frac{3}{2} x^{2})

\int \frac{x ^{3} - x + 2}{x - 1} d x

e^{- 2 t} cos^{2} (t)

\frac{\partial}{\partial y} (24 y^{2} - 12 x)

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