What is the integral from 0 to 4 of 4/(2x+1) ?

The integral from 0 to 4 of 4/(2x+1) is 4ln(3)

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integral from 0 to 4 of 4/(2x+1)

\int_{0}^{4} \frac{4}{2 x + 1} d x

4 ln (3)

Decimal Notation

4.39444 \dots

Solution steps

\int_{0}^{4} \frac{4}{2 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_{0}^{4} \frac{1}{2 x + 1} d x

Apply u-substitution

: \int_{1}^{9} \frac{1}{2 u} d u

= 4 \cdot \int_{1}^{9} \frac{1}{2 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}{2} \cdot \int_{1}^{9} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{2} [ln ∣ u ∣]_{1}^{9}

Simplify

4 \cdot \frac{1}{2} [ln ∣ u ∣]_{1}^{9} : 2 [ln ∣ u ∣]_{1}^{9}

= 2 [ln ∣ u ∣]_{1}^{9}

Compute the boundaries:

2 ln (3)

= 2 \cdot 2 ln (3)

Simplify

= 4 ln (3)

\frac{\partial}{\partial y} (e^{2 x y} cos (x))

\int 2 t^{2} d t

cos (3 t + 5) e^{2 t}

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

What is the integral from 0 to 4 of 4/(2x+1) ?
The integral from 0 to 4 of 4/(2x+1) is 4ln(3)