What is the integral of 4te^{7t^2} ?

The integral of 4te^{7t^2} is 2/7 e^{7t^2}+C

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integral of 4te^{7t^2}

\int 4 t e^{7 t^{2}} d t

\frac{2}{7} e^{7 t^{2}} + C

Solution steps

\int 4 t e^{7 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^{7 t^{2}} d t

Apply u-substitution

: \int \frac{e ^{u}}{14} d u

= 4 \cdot \int \frac{e ^{u}}{14} d u

Take the constant out:

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

= 4 \cdot \frac{1}{14} \cdot \int e^{u} d u

Use the common integral:

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

= 4 \cdot \frac{1}{14} e^{u}

Substitute back

u = 7 t^{2}

= 4 \cdot \frac{1}{14} e^{7 t^{2}}

Simplify

4 \cdot \frac{1}{14} e^{7 t^{2}} : \frac{2}{7} e^{7 t^{2}}

= \frac{2}{7} e^{7 t^{2}}

Add a constant to the solution

= \frac{2}{7} e^{7 t^{2}} + C

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What is the integral of 4te^{7t^2} ?
The integral of 4te^{7t^2} is 2/7 e^{7t^2}+C