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

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

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

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

2 e + \frac{e ^{2} - 1}{2} - 1

Decimal Notation

7.63109 \dots

Solution steps

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

Apply u-substitution

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

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

Expand

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

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

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

\int_{1}^{e} 2 d u = 2 e - 2

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

= \frac{e ^{2} - 1}{2} + 2 e - 2 + 1

Simplify

= 2 e + \frac{e ^{2} - 1}{2} - 1

\int 3 - x x^{2} d x

\int \frac{( x ^{3} - 2 x )}{x} d x

\frac{9}{( s ^{2} + 4 s + 9 )}

y = 8 x^{4}, at x = 2

f (x) = (x^{3} - 4 x)^{8}, at x = 2

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