What is the integral of x4^{x^2} ?

The integral of x4^{x^2} is (4^{x^2-1})/(ln(2))+C

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integral of x4^{x^2}

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

\frac{4 ^{x^{2} - 1}}{ln ( 2 )} + C

Solution steps

\int x \cdot 4^{x^{2}} d x

Apply u-substitution

: \int \frac{u}{ln ( 2 ) \cdot 4 ^{\frac{l n ( u ) + 2 l n ( 2 )}{2 l n ( 2 )}}} d u

= \int \frac{u}{ln ( 2 ) \cdot 4 ^{\frac{l n ( u ) + 2 l n ( 2 )}{2 l n ( 2 )}}} d u

Take the constant out:

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

= \frac{1}{ln ( 2 )} \cdot \int \frac{u}{4 ^{\frac{l n ( u ) + 2 l n ( 2 )}{2 l n ( 2 )}}} d u

Apply Integration By Parts

: \frac{u ^{2}}{2 ^{\frac{l n ( 4 u ) + l n ( 2 )}{l n ( 2 )}}} - \int - \frac{1}{8} d u

= \frac{1}{ln ( 2 )} (\frac{u ^{2}}{2 ^{\frac{l n ( 4 u ) + l n ( 2 )}{l n ( 2 )}}} - \int - \frac{1}{8} d u)

\int - \frac{1}{8} d u = - \frac{1}{8} u

= \frac{1}{ln ( 2 )} (\frac{u ^{2}}{2 ^{\frac{l n ( 4 u ) + l n ( 2 )}{l n ( 2 )}}} - (- \frac{1}{8} u))

Substitute back

u = 4^{x^{2}}

= \frac{1}{ln ( 2 )} \frac{( 4 ^{x^{2}} ) ^{2}}{2 ^{\frac{l n ( 4 \cdot 4 ^{x^{2}} ) + l n ( 2 )}{l n ( 2 )}}} - (- \frac{1}{8} \cdot 4^{x^{2}})

Simplify

\frac{1}{ln ( 2 )} \frac{( 4 ^{x^{2}} ) ^{2}}{2 ^{\frac{l n ( 4 \cdot 4 ^{x^{2}} ) + l n ( 2 )}{l n ( 2 )}}} - (- \frac{1}{8} \cdot 4^{x^{2}}) : \frac{4 ^{x^{2} - 1}}{ln ( 2 )}

= \frac{4 ^{x^{2} - 1}}{ln ( 2 )}

Add a constant to the solution

= \frac{4 ^{x^{2} - 1}}{ln ( 2 )} + C

\frac{d}{d t} (t e^{t})

g (t) = \frac{3 t ^{2} - 1}{2 t + 5}

\int 33 arctan (x) d x

x \to 0 - lim (\frac{e ^{\frac{2}{x}} \cdot sin ( π ( x ) )}{x + 2} + x)

x \to 0 lim (\frac{8 sin ( 0 )}{0})

What is the integral of x4^{x^2} ?
The integral of x4^{x^2} is (4^{x^2-1})/(ln(2))+C