What is the integral from 1 to ln(8) of 6x+e^x ?

The integral from 1 to ln(8) of 6x+e^x is 27ln^2(2)+5-e

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integral from 1 to ln(8) of 6x+e^x

\int_{1}^{l n (8)} 6 x + e^{x} d x

27 ln^{2} (2) + 5 - e

Decimal

15.25394 \dots

Solution steps

\int_{1}^{l n (8)} 6 x + e^{x} d x

Apply the Sum Rule:

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

= \int_{1}^{l n (8)} 6 x d x + \int_{1}^{l n (8)} e^{x} d x

\int_{1}^{l n (8)} 6 x d x = 3 (9 ln^{2} (2) - 1)

\int_{1}^{l n (8)} e^{x} d x = 8 - e

= 3 (9 ln^{2} (2) - 1) + 8 - e

Simplify

3 (9 ln^{2} (2) - 1) + 8 - e : 27 ln^{2} (2) + 5 - e

= 27 ln^{2} (2) + 5 - e

What is the integral from 1 to ln(8) of 6x+e^x ?
The integral from 1 to ln(8) of 6x+e^x is 27ln^2(2)+5-e