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

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

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integral of x*2^{x^2}

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

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

Solution steps

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

Apply u-substitution

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

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

Simplify

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

= \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

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

Apply exponent rule:

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

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

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

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

Substitute back

u = x^{2}

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

Simplify

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

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

Add a constant to the solution

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

f^{'} (x)

\int x sec^{2} (4 x) d x

f (x) = (x^{2} - 15)^{7}, at x = 4

x \frac{d y}{d x} + 2 y = 6 x^{2} y

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

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