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

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

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

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

- \frac{2}{e} + 3

Decimal Notation

2.26424 \dots

Solution steps

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

Apply u-substitution

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

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

Expand

\frac{u + 2}{u ^{2}} : \frac{1}{u} + \frac{2}{u ^{2}}

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

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}^{e} \frac{1}{u} d u + \int_{1}^{e} \frac{2}{u ^{2}} d u

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

\int_{1}^{e} \frac{2}{u ^{2}} d u = 2 (- \frac{1}{e} + 1)

= 1 + 2 (- \frac{1}{e} + 1)

Simplify

1 + 2 (- \frac{1}{e} + 1) : - \frac{2}{e} + 3

= - \frac{2}{e} + 3

\int_{0}^{3} 2 x - 3 d x

\int_{0}^{6} 3 x d x

\int_{1}^{3} \frac{x}{4 x ^{2}} d x

\int_{0}^{1} \frac{5}{7 + 7 x ^{2}} d x

\int_{0}^{4} (1 - x) d x

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