What is the integral of (4x^3+4x)e^{x^4+2x^2+1} ?

The integral of (4x^3+4x)e^{x^4+2x^2+1} is e^{1+x^4+2x^2}+C

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integral of (4x^3+4x)e^{x^4+2x^2+1}

\int (4 x^{3} + 4 x) e^{x^{4} + 2 x^{2} + 1} d x

e^{1 + x^{4} + 2 x^{2}} + C

Solution steps

\int (4 x^{3} + 4 x) e^{x^{4} + 2 x^{2} + 1} d x

(4 x^{3} + 4 x) e^{x^{4} + 2 x^{2} + 1} = 4 e^{1} (x^{3} + x) e^{x^{4} + 2 x^{2}}

= \int 4 e^{1} (x^{3} + x) e^{x^{4} + 2 x^{2}} d x

Take the constant out:

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

= 4 e^{1} \cdot \int (x^{3} + x) e^{x^{4} + 2 x^{2}} d x

Simplify

= 4 e \cdot \int (x^{3} + x) e^{x^{4} + 2 x^{2}} d x

Apply u-substitution

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

= 4 e \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

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

Use the common integral:

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

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

Substitute back

u = x^{4} + 2 x^{2}

= 4 e \frac{1}{4} e^{x^{4} + 2 x^{2}}

Simplify

4 e \frac{1}{4} e^{x^{4} + 2 x^{2}} : e^{1 + x^{4} + 2 x^{2}}

= e^{1 + x^{4} + 2 x^{2}}

Add a constant to the solution

= e^{1 + x^{4} + 2 x^{2}} + C

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What is the integral of (4x^3+4x)e^{x^4+2x^2+1} ?
The integral of (4x^3+4x)e^{x^4+2x^2+1} is e^{1+x^4+2x^2}+C