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Thread: Integral, please help find my mistake

  1. #1
    Junior Member TriForce's Avatar
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    Integral, please help find my mistake

    Ignore this post, found the mistake


    $$\int_{1}^{4}\frac{1}{x^{(3/2)}+x}dx = \int_{1}^{4}\frac{1}{\sqrt{x}*x+x}dx$$

    Subsitution:
    $$t=\sqrt{x}$$
    $$dt=\frac{1}{2t}dx$$

    $$\int_{1}^{2}\frac{2t}{t*t^2+t}dt = 2\int_{1}^{2}\frac{1}{t^2+1}dt = 2arctan(2) - 2arctan(1)$$

    Why is the above wrong?
    Last edited by topsquark; Apr 8th 2018 at 02:06 PM.
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  2. #2
    MHF Contributor
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    Re: Integral, please help find my mistake

    $x=t^2$ NOT $x=t$ so you end up with

    $\displaystyle \int_1^2 ~\dfrac{2t dt}{t^3 + t^2} = \int_1^2 ~\dfrac{2 dt}{t^2 + t}$

    it it however fairly trivial to continue using partial fractions
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  3. #3
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    Re: Integral, please help find my mistake

    Quote Originally Posted by TriForce View Post
    Ignore this post, found the mistake


    $$\int_{1}^{4}\frac{1}{x^(3/2)+x}dx = \int_{1}^{4}\frac{1}{\sqrt{x}*x+x}dx$$

    Subsitution:
    $$t=\sqrt{x}$$
    $$dt=\frac{1}{2t}dx$$

    $$\int_{1}^{2}\frac{2t}{t*t^2+t}dt = 2\int_{1}^{2}\frac{1}{t^2+1}dt = 2arctan(2) - 2arctan(1)$$

    Why is the above wrong?
    You replaced the x with t, but x=t^2
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