What is the integral from 0 to 2 of (e^{3y}-3)/(e^{3y)-9y+18} ?

The integral from 0 to 2 of (e^{3y}-3)/(e^{3y)-9y+18} is 1/3 (6-ln(19))

Popular >

integral from 0 to 2 of (e^{3y}-3)/(e^{3y)-9y+18}

\int_{0}^{2} \frac{e ^{3 y} - 3}{e ^{3 y} - 9 y + 18} d y

\frac{1}{3} (6 - ln (19))

Solution steps

\int_{0}^{2} \frac{e ^{3 y} - 3}{e ^{3 y} - 9 y + 18} d y

Apply u-substitution

: \int_{19}^{e^{6}} \frac{1}{3 u} d u

= \int_{19}^{e^{6}} \frac{1}{3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{19}^{e^{6}} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{19}^{e^{6}}

Compute the boundaries:

6 - ln (19)

= \frac{1}{3} (6 - ln (19))

\int_{0}^{1} x^{2} 1 + x^{3} d x

\int_{x}^{2 x} t^{3} d t

\int_{- 2}^{2} 8 d x

\int_{0}^{13} 26 x - x^{2} d x

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

What is the integral from 0 to 2 of (e^{3y}-3)/(e^{3y)-9y+18} ?
The integral from 0 to 2 of (e^{3y}-3)/(e^{3y)-9y+18} is 1/3 (6-ln(19))