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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int e^{v} \cdot 2 d v

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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