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

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

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

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

\frac{1}{3} e^{1 + x^{3} + 3 x} + C

Solution steps

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

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

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

Take the constant out:

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

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

Simplify

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

Apply u-substitution

: \int \frac{e ^{u}}{3} d u

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

Take the constant out:

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

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

Use the common integral:

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

= e \frac{1}{3} e^{u}

Substitute back

u = x^{3} + 3 x

= e \frac{1}{3} e^{x^{3} + 3 x}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

e e^{x^{3} + 3 x} = e^{1 + x^{3} + 3 x}

= \frac{1}{3} e^{1 + x^{3} + 3 x}

Add a constant to the solution

= \frac{1}{3} e^{1 + x^{3} + 3 x} + C

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