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

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

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integral from-1 to 1 of (4x-e^{-x})^2

\int_{- 1}^{1} (4 x - e^{- x})^{2} d x

\frac{96 e ^{2 W_{0} (- \frac{1}{4}) + 2} + 64 e ^{2 W_{0} (- \frac{1}{4}) + 3} + 3 e ^{2 W_{0} (- \frac{1}{4}) + 5} - 3 e ^{1 + 2 W_{0} (- \frac{1}{4})}}{6 e ^{2 W_{0} (- \frac{1}{4}) + 3}}

Solution steps

\int_{- 1}^{1} (4 x - e^{- x})^{2} d x

Multiply by the conjugate of

4 x - e^{- x} : \frac{16 x ^{2} - e ^{- 2 x}}{4 x + e ^{- x}}

= \int_{- 1}^{1} (\frac{16 x ^{2} - e ^{- 2 x}}{4 x + e ^{- x}})^{2} d x

Find undefined (singularity) points:

x = W_{0} (- \frac{1}{4})

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{- 1}^{W_{0} (- \frac{1}{4})} (\frac{16 x ^{2} - e ^{- 2 x}}{4 x + e ^{- x}})^{2} d x + \int_{W_{0} (- \frac{1}{4})}^{1} (\frac{16 x ^{2} - e ^{- 2 x}}{4 x + e ^{- x}})^{2} d x

\int_{- 1}^{W_{0} (- \frac{1}{4})} (\frac{16 x ^{2} - e ^{- 2 x}}{4 x + e ^{- x}})^{2} d x = \frac{16 ( W _{0} ( - \frac{1}{4} ) ^{3} + 1 )}{3} - 8 (- e^{- W_{0} (- \frac{1}{4})} W_{0} (- \frac{1}{4}) - e^{- W_{0} (- \frac{1}{4})}) + \frac{1}{2} (e^{2} - e^{- 2 W_{0} (- \frac{1}{4})})

\int_{W_{0} (- \frac{1}{4})}^{1} (\frac{16 x ^{2} - e ^{- 2 x}}{4 x + e ^{- x}})^{2} d x = \frac{16 ( 1 - W _{0} ( - \frac{1}{4} ) ^{3} )}{3} - 8 (- \frac{2}{e} + e^{- W_{0} (- \frac{1}{4})} W_{0} (- \frac{1}{4}) + e^{- W_{0} (- \frac{1}{4})}) + \frac{1}{2} (e^{- 2 W_{0} (- \frac{1}{4})} - \frac{1}{e ^{2}})

= \frac{16 ( W _{0} ( - \frac{1}{4} ) ^{3} + 1 )}{3} - 8 (- e^{- W_{0} (- \frac{1}{4})} W_{0} (- \frac{1}{4}) - e^{- W_{0} (- \frac{1}{4})}) + \frac{1}{2} (e^{2} - e^{- 2 W_{0} (- \frac{1}{4})}) + \frac{16 ( 1 - W _{0} ( - \frac{1}{4} ) ^{3} )}{3} - 8 (- \frac{2}{e} + e^{- W_{0} (- \frac{1}{4})} W_{0} (- \frac{1}{4}) + e^{- W_{0} (- \frac{1}{4})}) + \frac{1}{2} (e^{- 2 W_{0} (- \frac{1}{4})} - \frac{1}{e ^{2}})

Simplify

\frac{16 ( W _{0} ( - \frac{1}{4} ) ^{3} + 1 )}{3} - 8 (- e^{- W_{0} (- \frac{1}{4})} W_{0} (- \frac{1}{4}) - e^{- W_{0} (- \frac{1}{4})}) + \frac{1}{2} (e^{2} - e^{- 2 W_{0} (- \frac{1}{4})}) + \frac{16 ( 1 - W _{0} ( - \frac{1}{4} ) ^{3} )}{3} - 8 (- \frac{2}{e} + e^{- W_{0} (- \frac{1}{4})} W_{0} (- \frac{1}{4}) + e^{- W_{0} (- \frac{1}{4})}) + \frac{1}{2} (e^{- 2 W_{0} (- \frac{1}{4})} - \frac{1}{e ^{2}}) : \frac{96 e ^{2 W_{0} (- \frac{1}{4}) + 2} + 64 e ^{2 W_{0} (- \frac{1}{4}) + 3} + 3 e ^{2 W_{0} (- \frac{1}{4}) + 5} - 3 e ^{1 + 2 W_{0} (- \frac{1}{4})}}{6 e ^{2 W_{0} (- \frac{1}{4}) + 3}}

= \frac{96 e ^{2 W_{0} (- \frac{1}{4}) + 2} + 64 e ^{2 W_{0} (- \frac{1}{4}) + 3} + 3 e ^{2 W_{0} (- \frac{1}{4}) + 5} - 3 e ^{1 + 2 W_{0} (- \frac{1}{4})}}{6 e ^{2 W_{0} (- \frac{1}{4}) + 3}}

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