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

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

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

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

e^{t^{2} - 1} + C

Solution steps

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

Take the constant out:

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

= 2 \cdot \int t e^{t^{2} - 1} d t

Apply u-substitution

: \int \frac{e ^{u}}{2} d u

= 2 \cdot \int \frac{e ^{u}}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

= 2 \cdot \frac{1}{2} e^{u}

Substitute back

u = t^{2} - 1

= 2 \cdot \frac{1}{2} e^{t^{2} - 1}

Simplify

2 \cdot \frac{1}{2} e^{t^{2} - 1} : e^{t^{2} - 1}

= e^{t^{2} - 1}

Add a constant to the solution

= e^{t^{2} - 1} + C

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