What is the integral from 0 to 1 of (e^x+a^x) ?

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

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integral from 0 to 1 of (e^x+a^x)

\int_{0}^{1} (e^{x} + a^{x}) d x

e - 1 + \frac{a - 1}{ln ( a )}

Solution steps

\int_{0}^{1} e^{x} + a^{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_{0}^{1} e^{x} d x + \int_{0}^{1} a^{x} d x

\int_{0}^{1} e^{x} d x = e - 1

\int_{0}^{1} a^{x} d x = \frac{a - 1}{ln ( a )}

= e - 1 + \frac{a - 1}{ln ( a )}

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

\int_{0}^{x} \frac{x ^{4}}{1 - x ^{4}} d x

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

\int_{4}^{6} \frac{6}{x ^{4}} d x

\int_{1}^{3 x} ln (t) d t

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