Results 1 to 5 of 5

Math Help - integration question

  1. #1
    Newbie
    Joined
    Mar 2008
    Posts
    3

    integration question

    integrate 1 / (3-2x)^2. i dont understand how to integrate in this format, can someone please help me?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Mar 2008
    Posts
    2
    o__O

    Take t=3-2x
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hi,

    Or make appear the derivate of (3-2x) above, and you'll get u'(x)/uČ(x), and its integrate is -1/u
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member Nacho's Avatar
    Joined
    Mar 2008
    From
    Santiago, Chile
    Posts
    135
    Quote Originally Posted by -sonal- View Post
    integrate 1 / (3-2x)^2. i dont understand how to integrate in this format, can someone please help me?
    <br />
\begin{gathered}<br />
  \int {\frac{1}<br />
{{\left( {3 - 2x} \right)^2 }}dx}  \hfill \\<br />
   \hfill \\<br />
  u = 3 - 2x \Rightarrow du =  - 2dx \hfill \\ <br />
\end{gathered} <br />

    <br />
\int {\frac{1}<br />
{{\left( {3 - 2x} \right)^2 }}dx}  = \int {\frac{1}<br />
{{u^2 }} \cdot \frac{{ - 1}}<br />
{2}du}  = \frac{{ - 1}}<br />
{2}\int {u^{ - 2} du}  \Leftrightarrow \frac{{ - 1}}<br />
{2} \cdot \frac{{u^{ - 1} }}<br />
{{ - 1}} + C<br />

    <br />
\therefore \boxed{\int {\frac{1}<br />
{{\left( {3 - 2x} \right)^2 }}dx}  = \frac{1}<br />
{{2\left( {3 - 2x} \right)}} + C}<br />
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Nacho's answer is correct, but

    Quote Originally Posted by Nacho View Post

    \frac{{ - 1}}<br />
{2}\int {u^{ - 2} du}  \Leftrightarrow \frac{{ - 1}}<br />
{2} \cdot \frac{{u^{ - 1} }}<br />
{{ - 1}} + C<br />
    double implies it's applied when having equality, like 2a=2b\iff a=b. So in this case we require an equal sign.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. integration question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 30th 2011, 07:19 AM
  2. Replies: 6
    Last Post: July 21st 2010, 06:20 PM
  3. Integration Question
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 20th 2010, 02:59 AM
  4. help with integration question?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 19th 2010, 02:49 PM
  5. integration question no. 7
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 11th 2007, 08:40 AM

Search Tags


/mathhelpforum @mathhelpforum