What is the integral of (3^{2x}+3^{4x})^2 ?

The integral of (3^{2x}+3^{4x})^2 is (2*81^x+8*3^{6x-1}+6561^x)/(8ln(3))+C

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integral of (3^{2x}+3^{4x})^2

\int (3^{2 x} + 3^{4 x})^{2} d x

\frac{2 \cdot 8 1 ^{x} + 8 \cdot 3 ^{6 x - 1} + 656 1 ^{x}}{8 ln ( 3 )} + C

Solution steps

\int (3^{2 x} + 3^{4 x})^{2} d x

Apply u-substitution

: \int \frac{( 3 ^{u} + 9 ^{u} ) ^{2}}{2} d u

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int (3^{u} + 9^{u})^{2} d u

Apply Integration By Parts

: (3^{u} + 9^{u})^{2} u - \int 2 (3^{u} + 9^{u}) (3^{u} ln (3) + 2 ln (3) \cdot 9^{u}) u d u

= \frac{1}{2} ((3^{u} + 9^{u})^{2} u - \int 2 (3^{u} + 9^{u}) (3^{u} ln (3) + 2 ln (3) \cdot 9^{u}) u d u)

\int 2 (3^{u} + 9^{u}) (3^{u} ln (3) + 2 ln (3) \cdot 9^{u}) u d u = 9^{u} u - \frac{9 ^{u}}{2 ln ( 3 )} + 2 \cdot 2 7^{u} u - \frac{2 \cdot 3 ^{3 u - 1}}{ln ( 3 )} + 8 1^{u} u - \frac{8 1 ^{u}}{4 ln ( 3 )}

= \frac{1}{2} ((3^{u} + 9^{u})^{2} u - (9^{u} u - \frac{9 ^{u}}{2 ln ( 3 )} + 2 \cdot 2 7^{u} u - \frac{2 \cdot 3 ^{3 u - 1}}{ln ( 3 )} + 8 1^{u} u - \frac{8 1 ^{u}}{4 ln ( 3 )}))

Substitute back

u = 2 x

= \frac{1}{2} ((3^{2 x} + 9^{2 x})^{2} \cdot 2 x - (9^{2 x} \cdot 2 x - \frac{9 ^{2 x}}{2 ln ( 3 )} + 2 \cdot 2 7^{2 x} \cdot 2 x - \frac{2 \cdot 3 ^{3 \cdot 2 x - 1}}{ln ( 3 )} + 8 1^{2 x} \cdot 2 x - \frac{8 1 ^{2 x}}{4 ln ( 3 )}))

Simplify

\frac{1}{2} ((3^{2 x} + 9^{2 x})^{2} \cdot 2 x - (9^{2 x} \cdot 2 x - \frac{9 ^{2 x}}{2 ln ( 3 )} + 2 \cdot 2 7^{2 x} \cdot 2 x - \frac{2 \cdot 3 ^{3 \cdot 2 x - 1}}{ln ( 3 )} + 8 1^{2 x} \cdot 2 x - \frac{8 1 ^{2 x}}{4 ln ( 3 )})) : \frac{2 \cdot 8 1 ^{x} + 8 \cdot 3 ^{6 x - 1} + 656 1 ^{x}}{8 ln ( 3 )}

= \frac{2 \cdot 8 1 ^{x} + 8 \cdot 3 ^{6 x - 1} + 656 1 ^{x}}{8 ln ( 3 )}

Add a constant to the solution

= \frac{2 \cdot 8 1 ^{x} + 8 \cdot 3 ^{6 x - 1} + 656 1 ^{x}}{8 ln ( 3 )} + C

\frac{d}{d x} (17 x^{2} + 9)

x \to 10 lim (2 f (x))

2 y^{^{''}} + y^{^{'}} - y = x + 5, y (0) = 1, y^{^{'}} (0) = 0

\frac{d}{d x} (\frac{1}{5 x ^{4}})

\frac{d}{d x} (e^{- l n (x)})

What is the integral of (3^{2x}+3^{4x})^2 ?
The integral of (3^{2x}+3^{4x})^2 is (2*81^x+8*3^{6x-1}+6561^x)/(8ln(3))+C