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

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

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

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

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

Decimal Notation

11.25372 \dots

Solution steps

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

Apply u-substitution

: \int_{\frac{1}{e}}^{e} \frac{u ^{4} + 2 u ^{2} + 1}{u ^{3}} d u

= \int_{\frac{1}{e}}^{e} \frac{u ^{4} + 2 u ^{2} + 1}{u ^{3}} d u

Expand

\frac{u ^{4} + 2 u ^{2} + 1}{u ^{3}} : u + \frac{2}{u} + \frac{1}{u ^{3}}

= \int_{\frac{1}{e}}^{e} u + \frac{2}{u} + \frac{1}{u ^{3}} 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_{\frac{1}{e}}^{e} u d u + \int_{\frac{1}{e}}^{e} \frac{2}{u} d u + \int_{\frac{1}{e}}^{e} \frac{1}{u ^{3}} d u

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

\int_{\frac{1}{e}}^{e} \frac{2}{u} d u = 4

\int_{\frac{1}{e}}^{e} \frac{1}{u ^{3}} d u = - \frac{1}{2 e ^{2}} + \frac{e ^{2}}{2}

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

Simplify

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

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

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