What is the integral of (3^x+3^{-x})^2 ?

The integral of (3^x+3^{-x})^2 is (9^x)/(2ln(3))+2x-(9^{-x})/(2ln(3))+C

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integral of (3^x+3^{-x})^2

integral

(3^{x} + 3^{- x})^{2}

\frac{9 ^{x}}{2 ln ( 3 )} + 2 x - \frac{9 ^{- x}}{2 ln ( 3 )} + C

Solution steps

\int (3^{x} + 3^{- x})^{2} d x

Apply u-substitution

: \int - 9^{- u} - 2 - 9^{u} d u

= \int - 9^{- u} - 2 - 9^{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 9^{- u} d u - \int 2 d u - \int 9^{u} d u

\int 9^{- u} d u = - \frac{9 ^{- u}}{2 ln ( 3 )}

\int 2 d u = 2 u

\int 9^{u} d u = \frac{9 ^{u}}{2 ln ( 3 )}

= - (- \frac{9 ^{- u}}{2 ln ( 3 )}) - 2 u - \frac{9 ^{u}}{2 ln ( 3 )}

Substitute back

u = - x

= - (- \frac{9 ^{- (- x)}}{2 ln ( 3 )}) - 2 (- x) - \frac{9 ^{- x}}{2 ln ( 3 )}

Simplify

= \frac{9 ^{x}}{2 ln ( 3 )} + 2 x - \frac{9 ^{- x}}{2 ln ( 3 )}

Add a constant to the solution

= \frac{9 ^{x}}{2 ln ( 3 )} + 2 x - \frac{9 ^{- x}}{2 ln ( 3 )} + C

(\frac{1 - 1}{n})^{n - 1}

x^{2} ln (x + 1)

x (1 + x^{2})^{15}

\frac{x}{x ^{2} + y ^{2}}

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

What is the integral of (3^x+3^{-x})^2 ?
The integral of (3^x+3^{-x})^2 is (9^x)/(2ln(3))+2x-(9^{-x})/(2ln(3))+C