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

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

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

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

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

Solution steps

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

Apply u-substitution

= \int e^{u} d u

Use the common integral:

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

= e^{u}

Substitute back

u = x^{3} + x

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

Add a constant to the solution

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

\int \frac{1}{x - 2 x} d x

\int \frac{2 x + 3}{3 x ^{2} + 3 x - 1} d x

\int \frac{2 x ^{3} + 8 x}{x ^{2} + 4} d x

\int (e^{t} - e^{- t})^{2} d t

\int \frac{x ^{2} - 2 x + 6}{( x ^{2} + 2 ) ( x - 2 )} d x

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