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

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

Popular >

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

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

\frac{1}{2} (ln (3 e) - 1)

Decimal Notation

0.54930 \dots

Solution steps

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

Apply u-substitution

: \int_{e}^{3 e} \frac{1}{2 u} d u

= \int_{e}^{3 e} \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{e}^{3 e} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{e}^{3 e}

Compute the boundaries:

ln (3 e) - 1

= \frac{1}{2} (ln (3 e) - 1)

\int_{2}^{5} (t^{2} + 2 t) d t

\int_{0}^{2} y e^{- y^{2}} d y

\int_{0}^{\infty} sin^{2} (α) d α

\int_{- 1}^{2} ∣ x - 1 ∣ sin (x) d x

\int_{0}^{2} sin (x^{2}) x d x

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