1. ## integration question

integrate 1 / (3-2x)^2. i dont understand how to integrate in this format, can someone please help me?

2. o__O

Take t=3-2x

3. Hi,

Or make appear the derivate of (3-2x) above, and you'll get u'(x)/uČ(x), and its integrate is -1/u

4. Originally Posted by -sonal-
integrate 1 / (3-2x)^2. i dont understand how to integrate in this format, can someone please help me?
$\displaystyle \begin{gathered} \int {\frac{1} {{\left( {3 - 2x} \right)^2 }}dx} \hfill \\ \hfill \\ u = 3 - 2x \Rightarrow du = - 2dx \hfill \\ \end{gathered}$

$\displaystyle \int {\frac{1} {{\left( {3 - 2x} \right)^2 }}dx} = \int {\frac{1} {{u^2 }} \cdot \frac{{ - 1}} {2}du} = \frac{{ - 1}} {2}\int {u^{ - 2} du} \Leftrightarrow \frac{{ - 1}} {2} \cdot \frac{{u^{ - 1} }} {{ - 1}} + C$

$\displaystyle \therefore \boxed{\int {\frac{1} {{\left( {3 - 2x} \right)^2 }}dx} = \frac{1} {{2\left( {3 - 2x} \right)}} + C}$

5. Nacho's answer is correct, but

Originally Posted by Nacho

$\displaystyle \frac{{ - 1}} {2}\int {u^{ - 2} du} \Leftrightarrow \frac{{ - 1}} {2} \cdot \frac{{u^{ - 1} }} {{ - 1}} + C$
double implies it's applied when having equality, like $\displaystyle 2a=2b\iff a=b.$ So in this case we require an equal sign.