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?
$
\begin{gathered}
\int {\frac{1}
{{\left( {3 - 2x} \right)^2 }}dx} \hfill \\
\hfill \\
u = 3 - 2x \Rightarrow du = - 2dx \hfill \\
\end{gathered}
$

$
\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
$

$
\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

$\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 $2a=2b\iff a=b.$ So in this case we require an equal sign.