What is the integral of 16x^3e^{x^4+1} ?

The integral of 16x^3e^{x^4+1} is 4e^{x^4+1}+C

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integral of 16x^3e^{x^4+1}

\int 16 x^{3} e^{x^{4} + 1} d x

4 e^{x^{4} + 1} + C

Solution steps

\int 16 x^{3} e^{x^{4} + 1} d x

Take the constant out:

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

= 16 \cdot \int x^{3} e^{x^{4} + 1} d x

Apply u-substitution

: \int \frac{e ^{u}}{4} d u

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

Take the constant out:

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

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

Use the common integral:

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

= 16 \cdot \frac{1}{4} e^{u}

Substitute back

u = x^{4} + 1

= 16 \cdot \frac{1}{4} e^{x^{4} + 1}

Simplify

16 \cdot \frac{1}{4} e^{x^{4} + 1} : 4 e^{x^{4} + 1}

= 4 e^{x^{4} + 1}

Add a constant to the solution

= 4 e^{x^{4} + 1} + C

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What is the integral of 16x^3e^{x^4+1} ?
The integral of 16x^3e^{x^4+1} is 4e^{x^4+1}+C