What is the integral of x*9^{2x^2-8} ?

The integral of x*9^{2x^2-8} is (3^{4x^2-16})/(8ln(3))+C

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integral of x*9^{2x^2-8}

\int x \cdot 9^{2 x^{2} - 8} d x

\frac{3 ^{4 x^{2} - 16}}{8 ln ( 3 )} + C

Solution steps

\int x \cdot 9^{2 x^{2} - 8} d x

x 9^{2 x^{2} - 8} = 9^{- 8} x 9^{2 x^{2}}

= \int 9^{- 8} x \cdot 9^{2 x^{2}} d x

Take the constant out:

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

= 9^{- 8} \cdot \int x \cdot 9^{2 x^{2}} d x

9^{- 8} = \frac{1}{43046721}

= \frac{1}{43046721} \cdot \int x \cdot 9^{2 x^{2}} d x

Apply u-substitution

: \int \frac{8 1 ^{u}}{2} d u

= \frac{1}{43046721} \cdot \int \frac{8 1 ^{u}}{2} d u

Take the constant out:

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

= \frac{1}{43046721} \cdot \frac{1}{2} \cdot \int 8 1^{u} d u

\int a^{x} d x = \frac{a ^{x}}{ln a}

= \frac{1}{43046721} \cdot \frac{1}{2} \cdot \frac{8 1 ^{u}}{ln ( 81 )}

Substitute back

u = x^{2}

= \frac{1}{43046721} \cdot \frac{1}{2} \cdot \frac{8 1 ^{x^{2}}}{ln ( 81 )}

Simplify

\frac{1}{43046721} \cdot \frac{1}{2} \cdot \frac{8 1 ^{x^{2}}}{ln ( 81 )} : \frac{3 ^{4 x^{2} - 16}}{8 ln ( 3 )}

= \frac{3 ^{4 x^{2} - 16}}{8 ln ( 3 )}

Add a constant to the solution

= \frac{3 ^{4 x^{2} - 16}}{8 ln ( 3 )} + C

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

\int - \frac{1}{( t + 3 ) ^{2}} d t

\int cos^{3} (d) x d x

y = x^{2} + 3

x \to 0 + lim (x^{s i n (x)})

What is the integral of x*9^{2x^2-8} ?
The integral of x*9^{2x^2-8} is (3^{4x^2-16})/(8ln(3))+C