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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \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

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

Use the common integral:

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

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

Substitute back

u = t^{3} + 3 t

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

Add a constant to the solution

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

\int cos (\frac{π x}{4}) d x

f (x) = x^{4} - 4 x^{2} + 4

\frac{d}{d x} (sin (ln (tan (x))))

\int 6 t d t

y = (x + \frac{1}{x}) (x - \frac{1}{x})

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