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

The integral of (4x)/(4x+1) is x+1/4-1/4 ln|4x+1|+C

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

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

x + \frac{1}{4} - \frac{1}{4} ln ∣ 4 x + 1 ∣ + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{u - 1}{16 u} d u

= 4 \cdot \int \frac{u - 1}{16 u} d u

Take the constant out:

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

= 4 \cdot \frac{1}{16} \cdot \int \frac{u - 1}{u} d u

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

= 4 \cdot \frac{1}{16} \cdot \int 1 - \frac{1}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 \cdot \frac{1}{16} (\int 1 d u - \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= 4 \cdot \frac{1}{16} (u - ln ∣ u ∣)

Substitute back

u = 4 x + 1

= 4 \cdot \frac{1}{16} (4 x + 1 - ln ∣ 4 x + 1 ∣)

Simplify

4 \cdot \frac{1}{16} (4 x + 1 - ln ∣ 4 x + 1 ∣) : x + \frac{1}{4} - \frac{1}{4} ln ∣ 4 x + 1 ∣

= x + \frac{1}{4} - \frac{1}{4} ln ∣ 4 x + 1 ∣

Add a constant to the solution

= x + \frac{1}{4} - \frac{1}{4} ln ∣ 4 x + 1 ∣ + C

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What is the integral of (4x)/(4x+1) ?
The integral of (4x)/(4x+1) is x+1/4-1/4 ln|4x+1|+C