What is the integral from 1 to sqrt(3 of)x4^{x^2} ?

The integral from 1 to sqrt(3 of)x4^{x^2} is (15)/(ln(2))

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integral from 1 to sqrt(3 of)x4^{x^2}

\int_{1}^{3} x 4^{x^{2}} d x

\frac{15}{ln ( 2 )}

Decimal Notation

21.64042 \dots

Solution steps

\int_{1}^{3} x \cdot 4^{x^{2}} d x

Apply u-substitution

: \int_{4}^{64} \frac{u}{ln ( 2 ) \cdot 4 ^{\frac{l n ( u ) + 2 l n ( 2 )}{2 l n ( 2 )}}} d u

= \int_{4}^{64} \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_{4}^{64} \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]_{4}^{64}

= \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]_{4}^{64}

\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)]_{4}^{64}

Simplify

= \frac{1}{ln ( 2 )} [\frac{u ^{2}}{2 ^{\frac{l n ( 4 u ) + l n ( 2 )}{l n ( 2 )}}} + \frac{1}{8} u]_{4}^{64}

Compute the boundaries:

15

= \frac{1}{ln ( 2 )} \cdot 15

Simplify

= \frac{15}{ln ( 2 )}

\int \frac{1 - x ^{2}}{x ^{4}} arcsin (x) d x

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\frac{d ^{3} y}{d x ^{3}} + \frac{d ^{2} y}{d x ^{2}} - 2 y = 0

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\int \frac{1}{16 + y ^{2}} d y

What is the integral from 1 to sqrt(3 of)x4^{x^2} ?
The integral from 1 to sqrt(3 of)x4^{x^2} is (15)/(ln(2))