What is the integral of (8^x-8^{-x})^2 ?

The integral of (8^x-8^{-x})^2 is (2^{6x-1})/(3ln(2))-2x-(2^{-6x-1})/(3ln(2))+C

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integral of (8^x-8^{-x})^2

\int (8^{x} - 8^{- x})^{2} d x

\frac{2 ^{6 x - 1}}{3 ln ( 2 )} - 2 x - \frac{2 ^{- 6 x - 1}}{3 ln ( 2 )} + C

Solution steps

\int (8^{x} - 8^{- x})^{2} d x

Apply u-substitution

: \int - 6 4^{- u} + 2 - 6 4^{u} d u

= \int - 6 4^{- u} + 2 - 6 4^{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 6 4^{- u} d u + \int 2 d u - \int 6 4^{u} d u

\int 6 4^{- u} d u = - \frac{6 4 ^{- u}}{ln ( 64 )}

\int 2 d u = 2 u

\int 6 4^{u} d u = \frac{6 4 ^{u}}{ln ( 64 )}

= - (- \frac{6 4 ^{- u}}{ln ( 64 )}) + 2 u - \frac{6 4 ^{u}}{ln ( 64 )}

Substitute back

u = - x

= - (- \frac{6 4 ^{- (- x)}}{ln ( 64 )}) + 2 (- x) - \frac{6 4 ^{- x}}{ln ( 64 )}

Simplify

- (- \frac{6 4 ^{- (- x)}}{ln ( 64 )}) + 2 (- x) - \frac{6 4 ^{- x}}{ln ( 64 )} : \frac{2 ^{6 x - 1}}{3 ln ( 2 )} - 2 x - \frac{2 ^{- 6 x - 1}}{3 ln ( 2 )}

= \frac{2 ^{6 x - 1}}{3 ln ( 2 )} - 2 x - \frac{2 ^{- 6 x - 1}}{3 ln ( 2 )}

Add a constant to the solution

= \frac{2 ^{6 x - 1}}{3 ln ( 2 )} - 2 x - \frac{2 ^{- 6 x - 1}}{3 ln ( 2 )} + C

\frac{\partial}{\partial t} (y - e^{t})

\frac{d}{d x} ((2 cos (2 x) + sin (3 x))^{2})

e^{X}

\int y^{2} x d y

x y y^{^{'}} = 4 x^{2} + y^{2}

What is the integral of (8^x-8^{-x})^2 ?
The integral of (8^x-8^{-x})^2 is (2^{6x-1})/(3ln(2))-2x-(2^{-6x-1})/(3ln(2))+C