What is the integral of 1/(4x+6sqrt(x)) ?

The integral of 1/(4x+6sqrt(x)) is 1/2 ln|2sqrt(x)+3|+C

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integral of 1/(4x+6sqrt(x))

\int \frac{1}{4 x + 6 x} d x

\frac{1}{2} ln 2 x + 3 + C

Solution steps

\int \frac{1}{4 x + 6 x} d x

Apply u-substitution

: \int \frac{1}{2 u + 3} d u

= \int \frac{1}{2 u + 3} d u

Apply u-substitution

: \int \frac{1}{2 v} d v

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} ln ∣ v ∣

Substitute back

= \frac{1}{2} ln 2 x + 3

Add a constant to the solution

= \frac{1}{2} ln 2 x + 3 + C

n = 1 \sum \infty \frac{1}{2 n - 1}

x \frac{d y}{d x} = \frac{x ^{3}}{y} - \frac{5 x ^{5}}{y} - \frac{1}{x y}

f (x) = \frac{1}{x}, at x = 3

x \to + 0 lim ((1 + 6 x)^{\frac{4}{x}})

\frac{d}{d x} (- 2 e^{x} sin (x))

What is the integral of 1/(4x+6sqrt(x)) ?
The integral of 1/(4x+6sqrt(x)) is 1/2 ln|2sqrt(x)+3|+C