$\displaystyle \int_{1}^4 3/(2x + 1)^2\, dx$

2. Originally Posted by foodmmm
$\displaystyle \int_{1}^4 3/(2x + 1)^2\, dx$
Let $\displaystyle u=2x+1\Rightarrow du=2\,dx.$ Substitute.

3. Pull out the 3 (because it's a constant and you can do that ), and then try u-substitution with u being (2x + 1)

4. Originally Posted by foodmmm
$\displaystyle \int_{1}^4 3/(2x + 1)^2\, dx$
Notice that this is
$\displaystyle \int_1^4{3(2x + 1)^{-2}\,dx}$

$\displaystyle = 3\int_1^4{(2x + 1)^{-2}\,dx}$

$\displaystyle = \frac{3}{2}\int_1^4{2(2x + 1)^{-2}\,dx}$.

You would solve this using a u-substitution.

Let $\displaystyle u = 2x + 1$ so that $\displaystyle \frac{du}{dx} = 2$

So the integral becomes

$\displaystyle \frac{3}{2}\int_{x = 1}^{x = 4}{u^{-2}\,\frac{du}{dx}\,dx}$

$\displaystyle = \frac{3}{2}\int_{x = 1}^{x = 4}{u^{-2}\,du}$

$\displaystyle = \frac{3}{2}\left[-u^{-1}\right]_{x = 1}^{x = 4}$

$\displaystyle = \frac{3}{2}\left[-\frac{1}{2x + 1}\right]_1^4$

$\displaystyle = \frac{3}{2}\left[\left(-\frac{1}{9}\right) - \left(-\frac{1}{3}\right)\right]$

$\displaystyle = \frac{3}{2}\left(\frac{1}{3} - \frac{1}{9}\right)$

$\displaystyle = \frac{3}{2}\left(\frac{2}{9}\right)$

$\displaystyle = \frac{1}{3}$.

5. thank you. is there any way to do it without substitution? i don't think we are supposed to use it.

6. Originally Posted by foodmmm
thank you. is there any way to do it without substitution? i don't think we are supposed to use it.
Yes. Know that $\displaystyle \int \! (ax + b)^n \, dx = \frac{1}{a} \cdot \frac{1}{n+1} (ax + b)^{n+1}$, $\displaystyle n \neq -1$ (where I have omitted the arbitrary constant).