What is the integral from 0 to 4 of 5/(2x+1) ?

The integral from 0 to 4 of 5/(2x+1) is 5ln(3)

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integral from 0 to 4 of 5/(2x+1)

\int_{0}^{4} \frac{5}{2 x + 1} d x

5 ln (3)

Decimal

5.49306 \dots

Solution steps

\int_{0}^{4} \frac{5}{2 x + 1} d x

Take the constant out:

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

= 5 \cdot \int_{0}^{4} \frac{1}{2 x + 1} d x

Apply u-substitution

= 5 \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

= 5 \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 ∣)

= 5 \cdot \frac{1}{2} [ln ∣ u ∣]_{1}^{9}

Simplify

5 \cdot \frac{1}{2} [ln ∣ u ∣]_{1}^{9} : \frac{5}{2} [ln ∣ u ∣]_{1}^{9}

= \frac{5}{2} [ln ∣ u ∣]_{1}^{9}

Compute the boundaries:

2 ln (3)

= \frac{5}{2} \cdot 2 ln (3)

Simplify

= 5 ln (3)

d er i v a t i v e 3^{t}

ma c l a u r in x \cdot cos (2 x)

s im pl i f y lo g_{5} (x^{2} - 1)

t an g e n t y = (2)^{x}

x \to \infty lim (7)

What is the integral from 0 to 4 of 5/(2x+1) ?
The integral from 0 to 4 of 5/(2x+1) is 5ln(3)