Results 1 to 4 of 4

Thread: Integral 01

  1. #1
    Super Member
    Joined
    Jun 2008
    Posts
    829

    Integral 01

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



    Answer:
    $\displaystyle \frac{(2x^2-1)(1+x^2)\frac{1}{2}}{3x^3}$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2008
    Posts
    792
    $\displaystyle \int \frac{dx}{x^4\sqrt{1+x^2}}$

    You can use trigonometric subtitution.

    Let:
    $\displaystyle x = \tan{\theta}$

    $\displaystyle dx = \sec^2{\theta} d\theta$


    $\displaystyle \int \frac{\sec^2{\theta}}{(\tan{\theta})^4\sqrt{1+\tan ^2{\theta}}} d\theta$

    $\displaystyle \int \frac{\sec^2{\theta}}{(\tan{\theta})^4\sec{\theta} } d\theta$

    $\displaystyle \int \frac{\sec{\theta}}{(\tan{\theta})^4} d\theta$

    $\displaystyle \int \frac{1}{\cos{\theta}} \cdot \frac{cos^4{\theta}}{\sin^4{\theta}} d\theta$

    $\displaystyle \int \frac{cos^3{\theta}}{\sin^4{\theta}} d\theta$

    $\displaystyle \int \frac{(1-sin^2{\theta})\cos{\theta}}{\sin^4{\theta}} d\theta$

    Let
    $\displaystyle u = \sin{\theta}$

    $\displaystyle du = \cos{\theta} d\theta$

    $\displaystyle \int \frac{1-u^2}{u^4} du$

    Can you continue?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,656
    Thanks
    14
    Think in a reciprocal substitution, like $\displaystyle u=\frac1x,$ the integral equals $\displaystyle -\int{\frac{u^{3}}{\sqrt{1+u^{2}}}\,du}.$ Now put $\displaystyle z^2=1+u^2.$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jun 2008
    Posts
    829
    Look my solution


    $\displaystyle \int \frac{dx}{x^4\sqrt{1+x^2}} = \int x^(-4) (1+x^2)^-\frac{1}{2}dx$

    $\displaystyle 1+x^2=z^2$
    $\displaystyle x^2=z^2-1$
    $\displaystyle dx= \frac{zdz}{\sqrt{z^2-1}}$


    $\displaystyle \int (z^2+1)^(-2) (z^2)^\frac{1}{2}\frac{z}{\sqrt{z^2-1}}$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Aug 31st 2010, 07:38 AM
  2. Replies: 1
    Last Post: Jun 2nd 2010, 02:25 AM
  3. Replies: 0
    Last Post: May 9th 2010, 01:52 PM
  4. Replies: 0
    Last Post: Sep 10th 2008, 07:53 PM
  5. Replies: 6
    Last Post: May 18th 2008, 06:37 AM

Search Tags


/mathhelpforum @mathhelpforum