What is the integral of x^6e^{(x^7-4)} ?

The integral of x^6e^{(x^7-4)} is 1/7 e^{x^7-4}+C

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integral of x^6e^{(x^7-4)}

\int x^{6} e^{(x^{7} - 4)} d x

\frac{1}{7} e^{x^{7} - 4} + C

Solution steps

\int x^{6} e^{x^{7} - 4} d x

Apply u-substitution

: \int \frac{e ^{u^{\frac{7}{6}} - 4} u ^{\frac{1}{6}}}{6} d u

= \int \frac{e ^{u^{\frac{7}{6}} - 4} u ^{\frac{1}{6}}}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int e^{u^{\frac{7}{6}} - 4} u^{\frac{1}{6}} d u

Apply u-substitution

: \int \frac{6 e ^{v - 4}}{7} d v

= \frac{1}{6} \cdot \int \frac{6 e ^{v - 4}}{7} d v

Take the constant out:

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

= \frac{1}{6} \cdot \frac{6}{7} \cdot \int e^{v - 4} d v

e^{v - 4} = e^{- 4} e^{v}

= \frac{1}{6} \cdot \frac{6}{7} \cdot \int e^{- 4} e^{v} d v

Take the constant out:

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

= \frac{1}{6} \cdot \frac{6}{7} e^{- 4} \cdot \int e^{v} d v

Apply exponent rule:

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

= \frac{1}{6} \cdot \frac{6}{7} \cdot \frac{1}{e ^{4}} \cdot \int e^{v} d v

Use the common integral:

\int e^{v} d v = e^{v}

= \frac{1}{6} \cdot \frac{6}{7} \cdot \frac{1}{e ^{4}} e^{v}

Substitute back

= \frac{1}{6} \cdot \frac{6}{7} \cdot \frac{1}{e ^{4}} e^{(x^{6})^{\frac{7}{6}}}

Simplify

\frac{1}{6} \cdot \frac{6}{7} \cdot \frac{1}{e ^{4}} e^{(x^{6})^{\frac{7}{6}}} : \frac{1}{7} e^{x^{7} - 4}

= \frac{1}{7} e^{x^{7} - 4}

Add a constant to the solution

= \frac{1}{7} e^{x^{7} - 4} + C

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What is the integral of x^6e^{(x^7-4)} ?
The integral of x^6e^{(x^7-4)} is 1/7 e^{x^7-4}+C