What is the integral of 2/(4x+7sqrt(x)) ?

The integral of 2/(4x+7sqrt(x)) is ln|4sqrt(x)+7|+C

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integral of 2/(4x+7sqrt(x))

\int \frac{2}{4 x + 7 x} d x

ln 4 x + 7 + C

Solution steps

\int \frac{2}{4 x + 7 x} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{4 x + 7 x} d x

Apply u-substitution

: \int \frac{2}{4 u + 7} d u

= 2 \cdot \int \frac{2}{4 u + 7} d u

Take the constant out:

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

= 2 \cdot 2 \cdot \int \frac{1}{4 u + 7} d u

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 2 \cdot 2 \cdot \frac{1}{4} ln ∣ v ∣

Substitute back

= 2 \cdot 2 \cdot \frac{1}{4} ln 4 x + 7

Simplify

2 \cdot 2 \cdot \frac{1}{4} ln 4 x + 7 : ln 4 x + 7

= ln 4 x + 7

Add a constant to the solution

= ln 4 x + 7 + C

\frac{d y}{d t} = y^{2} + k

\int \frac{e ^{x}}{1 - e ^{x}} d x

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

\frac{1}{s ^{2} + 3 s}

x \to π - lim ((x + π) csc (x))

What is the integral of 2/(4x+7sqrt(x)) ?
The integral of 2/(4x+7sqrt(x)) is ln|4sqrt(x)+7|+C