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

The integral of te^{-7t^2} is -1/14 e^{-7t^2}+C

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integral of te^{-7t^2}

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

- \frac{1}{14} e^{- 7 t^{2}} + C

Solution steps

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

Apply u-substitution

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

= \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

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

Use the common integral:

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

= - \frac{1}{14} e^{u}

Substitute back

u = - 7 t^{2}

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

Add a constant to the solution

= - \frac{1}{14} e^{- 7 t^{2}} + C

e^{\frac{x}{y}}

\int \frac{1}{( 16 x ^{2} + 1 ) ^{\frac{3}{2}}} d x

\int 5 (\frac{sin ( 2 x )}{sin ( x )}) d x

\int \frac{1}{( e ^{x} + 1 ) ^{3} e ^{- x}} d x

t^{2} y^{^{''}} + 3 t y^{^{'}} + y = 0

What is the integral of te^{-7t^2} ?
The integral of te^{-7t^2} is -1/14 e^{-7t^2}+C