# Thread: simple integration by parts

1. ## simple integration by parts

How do you integrate (re^r/2)dr ?

Here is what I have so far, I think I am making a mistake with the integration of e^r/2.

(re^r/2)dr u = r dV = e^r/2dr
du = dr V = (1/2)e^r/2

uV - (integral of) Vdu

so:

(1/2)re^r/2 - (integral of)(1/2)e^r/2 dr

(1/2)re^r/2 - (1/4)e^r/2 + c

2. Originally Posted by mech.engineer.major
How do you integrate (re^r/2)dr ?

Here is what I have so far, I think I am making a mistake with the integration of e^r/2.

(re^r/2)dr u = r dV = e^r/2dr
du = dr V = (1/2)e^r/2

uV - (integral of) Vdu

so:

(1/2)re^r/2 - (integral of)(1/2)e^r/2 dr

(1/2)re^r/2 - (1/4)e^r/2 + c
Is the question $\int{re^{\frac{r}{2}}\,dr}$ or $\int{\frac{re^r}{2}\,dr}$?

3. the first of yours

4. To find $\int{re^{\frac{r}{2}}\,dr}$

Remember that

$\int{u\,\frac{dv}{dx}\,dx} = uv - \int{v\,\frac{du}{dx}\,dx}$

Let $u = r$ so that $\frac{du}{dr} = 1$.

Let $\frac{dv}{dr} = e^{\frac{r}{2}}$ so that $v = 2e^{\frac{r}{2}}$.

Therefore

$\int{re^{\frac{r}{2}}\,dr} = 2re^{\frac{r}{2}} - \int{2e^{\frac{r}{2}}\,dr}$

$= 2re^{\frac{r}{2}} - 4e^{\frac{r}{2}} + C$.

5. yeah thats what I thought I did wrong. I needed to divide e^(r/2) by the derivative of the exponent to get 2e^(r/2) instead of multiplying to get (1/2)e^(r/2).

thanks for your help