What is the integral of 3x*2^{x^2} ?

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

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integral of 3x*2^{x^2}

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int 2^{u - 1} d u

= 3 \cdot \int 2^{u - 1} d u

Simplify

2^{u - 1} : 2^{u} \cdot 2^{- 1}

= 3 \cdot \int 2^{u} \cdot 2^{- 1} d u

Take the constant out:

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

= 3 \cdot 2^{- 1} \cdot \int 2^{u} d u

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= 3 \cdot \frac{1}{2} \cdot \int 2^{u} d u

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

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

Substitute back

u = x^{2}

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

Simplify

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

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

Add a constant to the solution

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

x \to 0 lim (\frac{sin ( x )}{3 x})

x^{3} + y^{3} = 7 x y, (2, - 4)

y = \frac{1}{x ^{2}}, (- 1, 1)

\int_{\frac{1}{5}}^{7} 12 x ln (5 x) d x

\int \frac{- 1}{1 + x ^{2}} d x

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