What is the integral of (9x^2+4x+1)e^{3x^3+2x^2+x} ?

The integral of (9x^2+4x+1)e^{3x^3+2x^2+x} is e^{3x^3+2x^2+x}+C

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integral of (9x^2+4x+1)e^{3x^3+2x^2+x}

\int (9 x^{2} + 4 x + 1) e^{3 x^{3} + 2 x^{2} + x} d x

e^{3 x^{3} + 2 x^{2} + x} + C

Solution steps

\int (9 x^{2} + 4 x + 1) e^{3 x^{3} + 2 x^{2} + x} d x

Apply u-substitution

: \int e^{u} d u

= \int e^{u} d u

Use the common integral:

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

= e^{u}

Substitute back

u = 3 x^{3} + 2 x^{2} + x

= e^{3 x^{3} + 2 x^{2} + x}

Add a constant to the solution

= e^{3 x^{3} + 2 x^{2} + x} + C

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What is the integral of (9x^2+4x+1)e^{3x^3+2x^2+x} ?
The integral of (9x^2+4x+1)e^{3x^3+2x^2+x} is e^{3x^3+2x^2+x}+C