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

The integral from 0 to 1 of (e^x+e^{-x})^2 is 2((e^4-1)/(4e^2)+1)

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

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

2 (\frac{e ^{4} - 1}{4 e ^{2}} + 1)

Solution steps

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

Rewrite using trig identities

= \int_{0}^{1} (2 cosh (x))^{2} d x

Simplify

(2 cosh (x))^{2} : 4 cosh^{2} (x)

= \int_{0}^{1} 4 cosh^{2} (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int_{0}^{1} cosh^{2} (x) d x

Rewrite using trig identities

= 4 \cdot \int_{0}^{1} \frac{1 + cosh ( 2 x )}{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \frac{1}{2} \cdot \int_{0}^{1} 1 + cosh (2 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

= 4 \cdot \frac{1}{2} (\int_{0}^{1} 1 d x + \int_{0}^{1} cosh (2 x) d x)

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

\int_{0}^{1} cosh (2 x) d x = \frac{e ^{4} - 1}{4 e ^{2}}

= 4 \cdot \frac{1}{2} (1 + \frac{e ^{4} - 1}{4 e ^{2}})

Simplify

4 \cdot \frac{1}{2} (1 + \frac{e ^{4} - 1}{4 e ^{2}}) : 2 (\frac{e ^{4} - 1}{4 e ^{2}} + 1)

= 2 (\frac{e ^{4} - 1}{4 e ^{2}} + 1)

\int 2 cos (2 x) - 2 d x

\frac{d}{d x} (\frac{x - 7}{x + 2})

\int \frac{1}{12 - 4 x - x ^{2}} d x

x^{2} - 2, e^{x}, [- 1, 1]

\frac{( 4 x ^{3} )}{( 2 - 3 x ) ^{5}}

What is the integral from 0 to 1 of (e^x+e^{-x})^2 ?
The integral from 0 to 1 of (e^x+e^{-x})^2 is 2((e^4-1)/(4e^2)+1)