What is the integral from 0 to 1 of (8x^e+5e^x) ?

The integral from 0 to 1 of (8x^e+5e^x) is 8/(e+1)+5(e-1)

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integral from 0 to 1 of (8x^e+5e^x)

\int_{0}^{1} (8 x^{e} + 5 e^{x}) d x

\frac{8}{e + 1} + 5 (e - 1)

Decimal Notation

10.74294 \dots

Solution steps

\int_{0}^{1} 8 x^{e} + 5 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_{0}^{1} 8 x^{e} d x + \int_{0}^{1} 5 e^{x} d x

\int_{0}^{1} 8 x^{e} d x = \frac{8}{e + 1}

\int_{0}^{1} 5 e^{x} d x = 5 (e - 1)

= \frac{8}{e + 1} + 5 (e - 1)

\frac{d ^{2}}{d x ^{2}} (\frac{30 x}{( x ^{2} + 6 ) ^{2}})

7 y

\int \frac{1}{16 x ^{4} + 8 x ^{2} + 1} d x

\frac{\partial}{\partial x} (sin (k x))

What is the integral from 0 to 1 of (8x^e+5e^x) ?
The integral from 0 to 1 of (8x^e+5e^x) is 8/(e+1)+5(e-1)