What is the integral of te^{3t^2-1} ?

The integral of te^{3t^2-1} is 1/6 e^{3t^2-1}+C

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integral of te^{3t^2-1}

\int t e^{3 t^{2} - 1} d t

\frac{1}{6} e^{3 t^{2} - 1} + C

Solution steps

\int t e^{3 t^{2} - 1} d t

Apply u-substitution

= \int \frac{e ^{u}}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{6} e^{u}

Substitute back

u = 3 t^{2} - 1

= \frac{1}{6} e^{3 t^{2} - 1}

Add a constant to the solution

= \frac{1}{6} e^{3 t^{2} - 1} + C

What is the integral of te^{3t^2-1} ?
The integral of te^{3t^2-1} is 1/6 e^{3t^2-1}+C