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How do i find the antiderivative of this

  1. May 4, 2007 #1
    x/ square root sign(x^2 + 1)

    i did this x/ (x^2 + 1)^(1/2) simplified it

    then i did this (x^2/2)/ 3/2(x^2 + 1)^(3/2) antidirevitionated it

    then i cleaned it up x/ 3( x^2 + 1)^ 3

    is this right
     
    Last edited: May 4, 2007
  2. jcsd
  3. May 4, 2007 #2
    hmmm what did you do?

    consider the u-substitution of
    [tex]u=x^2[/tex]
     
  4. May 4, 2007 #3
    so i have to use the chain rule or the quotient rule when i find the antiderivative of this
     
  5. May 4, 2007 #4
    verify by differentiation that the formula x/square root (x^2 + 1) + C
    is correct

    this is what i have to do, but first i have to find the antiderivate of that function or polynomial, and then i have to just differentiate it which is easy
     
  6. May 5, 2007 #5
    Let u = x^2 + 1
    Then du = 2x*dx
    So x*dx/(x^2 + 1)^(1/2) = du/(2*u^(1/2))
     
  7. May 5, 2007 #6

    VietDao29

    User Avatar
    Homework Helper

    No, you should remember that there's no such thing as Chain rule, or Quotient Rule when finding an anti-derivative to one function.
    You should re-read your textbook again carefully, and slowly to fully understand what you have to do. In integration, we can make a u-substitution, a trig-substitution, Integration by Parts, blah, blah, blah,... but not chain rule, or quotient rule. They are just for differentiating.
    What you have done in your first post is not correct.
    [tex]\int \frac{f(x)}{g(x)} dx \neq \frac{\int f(x) dx}{\int g(x) dx}[/tex], they are not the same.

    -----------------------

    I'll give you an example, similar to the problem. But you really should read your text book first.

    Example 1:
    [tex]\int x \sqrt{x ^ 2 + 3} dx[/tex]

    You should notice the x in front of the square root, it differs from the derivative of (x2 + 3, i.e, the expression inside the square root) by a factor 2.

    [tex]\int x \sqrt{x ^ 2 + 3} dx = \int \sqrt{x ^ 2 + 3} x dx[/tex]
    Let [tex]u = x ^ 2 + 3 \Rightarrow du = (x ^ 2 + 3)'_x dx = 2x dx \Rightarrow x dx = \frac{du}{2}[/tex]
    The integral becomes:
    [tex]... = \int \sqrt{u} \frac{du}{2} = \frac{1}{2} \int \sqrt{u} du = \frac{1}{2} \int u ^ {\frac{1}{2}} du = \frac{1}{2} \times \frac{u ^ {\frac{3}{2}}}{\frac{3}{2}} + C[/tex]

    [tex]= \frac{1}{3} u ^ {\frac{3}{2}} + C = \frac{1}{3} \sqrt{u ^ 3} + C[/tex], now, change u back to x, we have:

    [tex]... = \frac{1}{3} \sqrt{(x ^ 2 + 3) ^ 3} + C[/tex]

    Example 2:
    [tex]\int x e ^ {3x ^ 2 + 5} dx[/tex]

    It's pretty similar to the previous one, let [tex]u = 3x ^ 2 + 5 \Rightarrow du = 6x dx \Rightarrow x dx = \frac{du}{6}[/tex]
    The integral becomes:
    [tex]\int e ^ u \frac{du}{6} = \frac{1}{6} \int e ^ u du = \frac{1}{6} e ^ u + C = \frac{1}{6} e ^ {3x ^ 2 + 5} + C[/tex]

    Now, you can just do the same to your problem. Can you go from here? :)
     
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