What is the integral of x^4*2^{x^5+7} ?

The integral of x^4*2^{x^5+7} is (2^{7+x^5})/(5ln(2))+C

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integral of x^4*2^{x^5+7}

\int x^{4} \cdot 2^{x^{5} + 7} d x

\frac{2 ^{7 + x^{5}}}{5 ln ( 2 )} + C

Solution steps

\int x^{4} \cdot 2^{x^{5} + 7} d x

x^{4} 2^{x^{5} + 7} = 2^{7} x^{4} 2^{x^{5}}

= \int 2^{7} x^{4} \cdot 2^{x^{5}} d x

Take the constant out:

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

= 2^{7} \cdot \int x^{4} \cdot 2^{x^{5}} d x

Simplify

= 128 \cdot \int x^{4} \cdot 2^{x^{5}} d x

Apply u-substitution

: \int 2^{u^{\frac{5}{4}} - 2} u^{\frac{1}{4}} d u

= 128 \cdot \int 2^{u^{\frac{5}{4}} - 2} u^{\frac{1}{4}} d u

Apply u-substitution

: \int \frac{2 ^{v}}{5} d v

= 128 \cdot \int \frac{2 ^{v}}{5} d v

Take the constant out:

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

= 128 \cdot \frac{1}{5} \cdot \int 2^{v} d v

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

= 128 \cdot \frac{1}{5} \cdot \frac{2 ^{v}}{ln ( 2 )}

Substitute back

= 128 \cdot \frac{1}{5} \cdot \frac{2 ^{(x^{4})^{\frac{5}{4}}}}{ln ( 2 )}

Simplify

128 \cdot \frac{1}{5} \cdot \frac{2 ^{(x^{4})^{\frac{5}{4}}}}{ln ( 2 )} : \frac{2 ^{7 + x^{5}}}{5 ln ( 2 )}

= \frac{2 ^{7 + x^{5}}}{5 ln ( 2 )}

Add a constant to the solution

= \frac{2 ^{7 + x^{5}}}{5 ln ( 2 )} + C

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What is the integral of x^4*2^{x^5+7} ?
The integral of x^4*2^{x^5+7} is (2^{7+x^5})/(5ln(2))+C