Results 1 to 8 of 8

Math Help - Stuck on several integral problems

  1. #1
    Boz
    Boz is offline
    Newbie
    Joined
    Nov 2010
    Posts
    17

    Stuck on several integral problems

    I'll appreciate any help on evaluating the following integrals:

    \int\frac{x+1}{x^2+2x+4}dx

    \int\frac{x}{x^2+2x-3}dx

    \int\frac{x+1}{x^2-2x-3}dx


    Thank you in advance
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,654
    Thanks
    1478
    1. \displaystyle \int{\frac{x+1}{x^2 + 2x + 4}\,dx} = \frac{1}{2}\int{\frac{2x+2}{x^2 + 2x + 4}\,dx}. Now make the substitution \displaystyle u=x^2 + 2x + 4.


    2. You will need to use a Partial Fraction decomposition.

    \displaystyle \frac{A}{x+3} + \frac{B}{x-1} = \frac{x}{(x+3)(x-1)}

    \displaystyle \frac{A(x-1) + B(x+3)}{(x+3)(x-1)} = \frac{x}{(x+3)(x-1)}

    \displaystyle A(x-1)+B(x+3) = x

    \displaystyle Ax-A+Bx+3B = x

    \displaystyle (A+B)x -A+3B= 1x + 0.


    So \displaystyle A+B = 1 and \displaystyle -A+3B=0.

    Solving simultaneously gives \displaystyle A = \frac{3}{4}, B = \frac{1}{4}.


    So \displaystyle \int{\frac{x}{x^2+2x-3}\,dx} = \frac{3}{4}\int{\frac{1}{x+3}\,dx} + \frac{1}{4}\int{\frac{1}{x-1}\,dx}. Go from here.


    3. Note that \displaystyle \frac{x+1}{x^2-2x - 3} = \frac{x+1}{(x - 3)(x + 1)} = \frac{1}{x-3}.

    So \displaystyle \int{\frac{x+1}{x^2-2x-3}\,dx} = \int{\frac{1}{x-3}\,dx}. Go from here.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,792
    Thanks
    1687
    Awards
    1
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Boz
    Boz is offline
    Newbie
    Joined
    Nov 2010
    Posts
    17
    Thank you so much for the quick response.It was very helpful.
    Here is what I did. Is it correct?

    Problem 1  \displaystyle x^2+2x+4=u, \frac{du}{dx}=2x+2. Hence, dx=\frac{du}{2(x+1)}
    \displaystyle Then<br />
\int  \frac{x+1}{u}\frac{du}{2(x+1)}=\int \frac{1}{2} \frac{1}{u}du=\frac<br />
{1}{2}\ln u =\frac{1}{2}\ln(x^2+2x+4)

    Problem 2 \displaystyle\int \frac{1}{4(x-1)}+\frac{3}{4(x+3)} dx
    \displaystyle\int\frac{1}{4}\frac{1}{x_{1}}+\int\f  rac{3}{4}\frac{1}{x_{2}}=\frac{\ln(x-1)}{4}+\frac{3ln(x+3)}{4}+C
     \displaystyle\frac{1}{4}\ln(x-1) + \frac{3}{4}\ln(x+3) + C<br />

    Problem 3 \displaystyle\int\frac{x+1}{(x+1)(x-3)} = \int\frac{1}{x-3} dx = \ln(x-3) + C<br />
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Boz
    Boz is offline
    Newbie
    Joined
    Nov 2010
    Posts
    17
    Quote Originally Posted by Plato View Post
    Thx about the recommendation. The site is great. You can watch the TED video of Wolfram if you haven't done it.It's very interesting.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,654
    Thanks
    1478
    Quote Originally Posted by Boz View Post
    Thank you so much for the quick response.It was very helpful.
    Here is what I did. Is it correct?

    Problem 1  \displaystyle x^2+2x+4=u, \frac{du}{dx}=2x+2. Hence, dx=\frac{du}{2(x+1)}
    \displaystyle Then<br />
\int  \frac{x+1}{u}\frac{du}{2(x+1)}=\int \frac{1}{2} \frac{1}{u}du=\frac<br />
{1}{2}\ln u =\frac{1}{2}\ln(x^2+2x+4)

    Problem 2 \displaystyle\int \frac{1}{4(x-1)}+\frac{3}{4(x+3)} dx
    \displaystyle\int\frac{1}{4}\frac{1}{x_{1}}+\int\f  rac{3}{4}\frac{1}{x_{2}}=\frac{\ln(x-1)}{4}+\frac{3ln(x+3)}{4}+C
     \displaystyle\frac{1}{4}\ln(x-1) + \frac{3}{4}\ln(x+3) + C<br />

    Problem 3 \displaystyle\int\frac{x+1}{(x+1)(x-3)} = \int\frac{1}{x-3} dx = \ln(x-3) + C<br />
    1. Correct.

    2. I don't know why you have decided midway to introduce \displaystyle x_1 and \displaystyle x_2, it's still a function of \displaystyle x.

    Also, recall that \displaystyle \int{\frac{1}{ax+b}\,dx} = \frac{1}{a}\ln{|ax+b|}. You can't leave off the modulus. You could in question 1 because \displaystyle x^2 + 2x + 4 is always positive...

    3. Again, you can't leave off the modulus.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Boz
    Boz is offline
    Newbie
    Joined
    Nov 2010
    Posts
    17
    Quote Originally Posted by Prove It View Post
    1. Correct.

    2. I don't know why you have decided midway to introduce \displaystyle x_1 and \displaystyle x_2, it's still a function of \displaystyle x.

    Also, recall that \displaystyle \int{\frac{1}{ax+b}\,dx} = \frac{1}{a}\ln{|ax+b|}. You can't leave off the modulus. You could in question 1 because \displaystyle x^2 + 2x + 4 is always positive...

    3. Again, you can't leave off the modulus.
    I put x_{1} just to make it easier for myself because I am still learning this integration technique.
    So, the answers are going to be :

    Pr.2 \frac{1}{4}ln|x-1|+\frac{3}{4}ln|x+3|+C
    Pr.3 ln|x-3|+C ??
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor harish21's Avatar
    Joined
    Feb 2010
    From
    Dirty South
    Posts
    1,036
    Thanks
    10
    The answers you have written for No. 2 and No. 3 are correct!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. stuck on some problems
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 5th 2010, 06:51 PM
  2. three problems I am stuck on
    Posted in the Number Theory Forum
    Replies: 6
    Last Post: March 5th 2009, 01:57 PM
  3. A Couple Integral Problems I'm Stuck On
    Posted in the Calculus Forum
    Replies: 3
    Last Post: January 25th 2009, 04:49 PM
  4. Stuck on a few problems
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: September 6th 2008, 08:29 PM
  5. A few Webwork Integral problems have me stuck.
    Posted in the Calculus Forum
    Replies: 13
    Last Post: May 30th 2007, 05:05 PM

Search Tags


/mathhelpforum @mathhelpforum