What is the integral of 4te^{t^2} ?

The integral of 4te^{t^2} is 2e^{t^2}+C

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integral of 4te^{t^2}

\int 4 t e^{t^{2}} d t

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

Solution steps

\int 4 t e^{t^{2}} d t

Take the constant out:

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

= 4 \cdot \int t e^{t^{2}} d t

Apply u-substitution

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

= 4 \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

= 4 \cdot \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

= 4 \cdot \frac{1}{2} e^{u}

Substitute back

u = t^{2}

= 4 \cdot \frac{1}{2} e^{t^{2}}

Simplify

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

= 2 e^{t^{2}}

Add a constant to the solution

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

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