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

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

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

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

2 ln x + 1 + C

Solution steps

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

Apply u-substitution

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

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

Add a constant to the solution

= 2 ln x + 1 + C

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

x \to - 1 lim (3 x^{2} - 2 x - 1)

f (x) = 3 - x^{2} + 2 x

f (x) = 8 x^{2} + 4 x - 6, at x = - 3

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

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