What is the integral of t/(e^{t^2)} ?

The integral of t/(e^{t^2)} is -1/2 e^{-t^2}+C

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integral of t/(e^{t^2)}

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

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

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{t}{e ^{t^{2}}} = (t) e^{- t^{2}}

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

Apply u-substitution

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

= \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

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

Use the common integral:

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

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

Substitute back

u = - t^{2}

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

Add a constant to the solution

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

\int_{0}^{\infty} x^{3} e^{- x} d x

\frac{d}{d x} (sech (x))

x^{3} \cdot e^{x}

x \to 1 lim (- \frac{2}{x ^{3}})

\int \frac{7 x ^{6}}{( 4 + x ^{7} ) ^{5}} d x

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