What is the integral of x^5e^{x^3-7} ?

The integral of x^5e^{x^3-7} is 1/3 e^{x^3-7}(x^3-1)+C

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integral of x^5e^{x^3-7}

\int x^{5} e^{x^{3} - 7} d x

\frac{1}{3} e^{x^{3} - 7} (x^{3} - 1) + C

Solution steps

\int x^{5} e^{x^{3} - 7} d x

x^{5} e^{x^{3} - 7} = e^{- 7} x^{5} e^{x^{3}}

= \int e^{- 7} x^{5} e^{x^{3}} d x

Take the constant out:

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

= e^{- 7} \cdot \int x^{5} e^{x^{3}} d x

Apply exponent rule:

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

= \frac{1}{e ^{7}} \cdot \int x^{5} e^{x^{3}} d x

Apply u-substitution

= \frac{1}{e ^{7}} \cdot \int \frac{e ^{u^{\frac{3}{5}}} u ^{\frac{1}{5}}}{5} d u

Take the constant out:

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

= \frac{1}{e ^{7}} \cdot \frac{1}{5} \cdot \int e^{u^{\frac{3}{5}}} u^{\frac{1}{5}} d u

Apply u-substitution

= \frac{1}{e ^{7}} \cdot \frac{1}{5} \cdot \int \frac{5 e ^{v} v}{3} d v

Take the constant out:

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

= \frac{1}{e ^{7}} \cdot \frac{1}{5} \cdot \frac{5}{3} \cdot \int e^{v} v d v

Apply Integration By Parts

= \frac{1}{e ^{7}} \cdot \frac{1}{5} \cdot \frac{5}{3} (e^{v} v - \int e^{v} d v)

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

= \frac{1}{e ^{7}} \cdot \frac{1}{5} \cdot \frac{5}{3} (e^{v} v - e^{v})

Substitute back

= \frac{1}{e ^{7}} \cdot \frac{1}{5} \cdot \frac{5}{3} (e^{(x^{5})^{\frac{3}{5}}} (x^{5})^{\frac{3}{5}} - e^{(x^{5})^{\frac{3}{5}}})

Simplify

\frac{1}{e ^{7}} \cdot \frac{1}{5} \cdot \frac{5}{3} (e^{(x^{5})^{\frac{3}{5}}} (x^{5})^{\frac{3}{5}} - e^{(x^{5})^{\frac{3}{5}}}) : \frac{1}{3} e^{x^{3} - 7} (x^{3} - 1)

= \frac{1}{3} e^{x^{3} - 7} (x^{3} - 1)

Add a constant to the solution

= \frac{1}{3} e^{x^{3} - 7} (x^{3} - 1) + C

What is the integral of x^5e^{x^3-7} ?
The integral of x^5e^{x^3-7} is 1/3 e^{x^3-7}(x^3-1)+C