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

The integral of te^{t/2} is 4(1/2 e^{t/2}t-e^{t/2})+C

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integral of te^{t/2}

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

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

Solution steps

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

Apply u-substitution

: \int 4 e^{u} u d u

= \int 4 e^{u} u d u

Take the constant out:

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

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

Apply Integration By Parts

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

= 4 (e^{u} u - \int e^{u} d u)

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

= 4 (e^{u} u - e^{u})

Substitute back

u = \frac{t}{2}

= 4 (e^{\frac{t}{2}} \frac{t}{2} - e^{\frac{t}{2}})

Simplify

4 (e^{\frac{t}{2}} \frac{t}{2} - e^{\frac{t}{2}}) : 4 (\frac{1}{2} e^{\frac{t}{2}} t - e^{\frac{t}{2}})

= 4 (\frac{1}{2} e^{\frac{t}{2}} t - e^{\frac{t}{2}})

Add a constant to the solution

= 4 (\frac{1}{2} e^{\frac{t}{2}} t - e^{\frac{t}{2}}) + C

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What is the integral of te^{t/2} ?
The integral of te^{t/2} is 4(1/2 e^{t/2}t-e^{t/2})+C