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

The integral of (e^x+e^{-x})^2 is 2x+sinh(2x)+C

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integral of (e^x+e^{-x})^2

integral

(e^{x} + e^{- x})^{2}

2 x + sinh (2 x) + C

Solution steps

\int (e^{x} + e^{- x})^{2} d x

Rewrite using trig identities

= \int (2 cosh (x))^{2} d x

Simplify

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

= \int 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 cosh^{2} (x) d x

Rewrite using trig identities

= 4 \cdot \int \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 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 1 d x + \int cosh (2 x) d x)

\int 1 d x = x

\int cosh (2 x) d x = \frac{1}{2} sinh (2 x)

= 4 \cdot \frac{1}{2} (x + \frac{1}{2} sinh (2 x))

Simplify

4 \cdot \frac{1}{2} (x + \frac{1}{2} sinh (2 x)) : 2 x + sinh (2 x)

= 2 x + sinh (2 x)

Add a constant to the solution

= 2 x + sinh (2 x) + C

\frac{1}{( e ^{x} + e ^{- x} )}

\frac{sin ^{2} ( x )}{( 1 + cos ( x ) ) ^{2}}

cos^{6} (x)

\frac{1}{a ^{x}}

\frac{1}{( x ^{2} + 2 )}

What is the integral of (e^x+e^{-x})^2 ?
The integral of (e^x+e^{-x})^2 is 2x+sinh(2x)+C