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

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

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

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

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

Solution steps

\int x \cdot 3^{- x^{2}} d x

Apply u-substitution

: \int - \frac{1}{2 ln ( 3 )} d u

= \int - \frac{1}{2 ln ( 3 )} d u

Integral of a constant:

\int a d x = a x

= (- \frac{1}{2 ln ( 3 )}) u

Substitute back

u = 3^{- x^{2}}

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

Simplify

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

Add a constant to the solution

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

x \to 1 lim (1.7496)

f (x) = 12 x

h \to 0 lim (4 - h)

y = sin (x) tan (x)

\frac{d}{d x} (3 cos^{4} (x))

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