What is the integral of 16x^7e^{2x^4} ?

The integral of 16x^7e^{2x^4} is 2e^{2x^4}x^4-e^{2x^4}+C

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integral of 16x^7e^{2x^4}

\int 16 x^{7} e^{2 x^{4}} d x

2 e^{2 x^{4}} x^{4} - e^{2 x^{4}} + C

Solution steps

\int 16 x^{7} e^{2 x^{4}} d x

Take the constant out:

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

= 16 \cdot \int x^{7} e^{2 x^{4}} d x

Apply u-substitution

= 16 \cdot \int \frac{e ^{2 u^{\frac{4}{7}}} u ^{\frac{1}{7}}}{7} d u

Take the constant out:

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

= 16 \cdot \frac{1}{7} \cdot \int e^{2 u^{\frac{4}{7}}} u^{\frac{1}{7}} d u

Apply u-substitution

= 16 \cdot \frac{1}{7} \cdot \int \frac{7 e ^{2 v} v}{4} d v

Take the constant out:

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

= 16 \cdot \frac{1}{7} \cdot \frac{7}{4} \cdot \int e^{2 v} v d v

Apply u-substitution

= 16 \cdot \frac{1}{7} \cdot \frac{7}{4} \cdot \int \frac{e ^{w} w}{4} d w

Take the constant out:

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

= 16 \cdot \frac{1}{7} \cdot \frac{7}{4} \cdot \frac{1}{4} \cdot \int e^{w} w d w

Apply Integration By Parts

= 16 \cdot \frac{1}{7} \cdot \frac{7}{4} \cdot \frac{1}{4} (e^{w} w - \int e^{w} d w)

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

= 16 \cdot \frac{1}{7} \cdot \frac{7}{4} \cdot \frac{1}{4} (e^{w} w - e^{w})

Substitute back

= 16 \cdot \frac{1}{7} \cdot \frac{7}{4} \cdot \frac{1}{4} (e^{2 (x^{7})^{\frac{4}{7}}} \cdot 2 (x^{7})^{\frac{4}{7}} - e^{2 (x^{7})^{\frac{4}{7}}})

Simplify

16 \cdot \frac{1}{7} \cdot \frac{7}{4} \cdot \frac{1}{4} (e^{2 (x^{7})^{\frac{4}{7}}} \cdot 2 (x^{7})^{\frac{4}{7}} - e^{2 (x^{7})^{\frac{4}{7}}}) : 2 e^{2 x^{4}} x^{4} - e^{2 x^{4}}

= 2 e^{2 x^{4}} x^{4} - e^{2 x^{4}}

Add a constant to the solution

= 2 e^{2 x^{4}} x^{4} - e^{2 x^{4}} + C

What is the integral of 16x^7e^{2x^4} ?
The integral of 16x^7e^{2x^4} is 2e^{2x^4}x^4-e^{2x^4}+C