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Math Help - Integrate In anyway possible.

  1. #1
    Junior Member EliteAndoy's Avatar
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    Integrate In anyway possible.

    Hi everyone! Today we are asked in class to solve an integral in anyway that we know. The given integral is \int \frac{x^4}{x-1}dx. She hints that it is possible to solve it using integration by parts, and I see that I might have to use tablature method to find the integral since I have to reduce x^4 to zero power. But I said, screw it! So I attempted to do it by doing somewhat partial fraction. So here's what I did:

    1)I did a long division so that:

    \frac{x^4}{x-1}=x^3+x^2+x+1+\frac{1}{x-1}

    2)Turns out I don't have to solve for a partial fraction so:

    \int \frac{x^4}{x-1}dx= \int x^3+x^2+x+1+\frac{1}{x-1}dx

    3)So that if we integrate it, it becomes:

    \int \frac{x^4}{x-1}dx=\frac{x^4}{4}+\frac{x^3}{3}+\frac{x^2}{2}+x+l  n\left |x-1\right |+c

    Is that solution good enough? Not really sure about it. Anyone got any ideas to evaluate the integral more efficiently? Thanks a lot in advance!
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  2. #2
    MHF Contributor
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    Re: Integrate In anyway possible.

    Quote Originally Posted by EliteAndoy View Post
    Hi everyone! Today we are asked in class to solve an integral in anyway that we know. The given integral is \int \frac{x^4}{x-1}dx. She hints that it is possible to solve it using integration by parts, and I see that I might have to use tablature method to find the integral since I have to reduce x^4 to zero power. But I said, screw it! So I attempted to do it by doing somewhat partial fraction. So here's what I did:

    1)I did a long division so that:

    \frac{x^4}{x-1}=x^3+x^2+x+1+\frac{1}{x-1}

    2)Turns out I don't have to solve for a partial fraction so:

    \int \frac{x^4}{x-1}dx= \int x^3+x^2+x+1+\frac{1}{x-1}dx

    3)So that if we integrate it, it becomes:

    \int \frac{x^4}{x-1}dx=\frac{x^4}{4}+\frac{x^3}{3}+\frac{x^2}{2}+x+l  n\left |x-1\right |+c

    Is that solution good enough? Not really sure about it. Anyone got any ideas to evaluate the integral more efficiently? Thanks a lot in advance!
    Let \displaystyle \begin{align*} u = x - 1 \implies du = dx \end{align*} and the integral becomes

    \displaystyle \begin{align*} \int{\frac{x^4}{x - 1}\,dx} &= \int{\frac{\left( u + 1 \right)^4}{u}\,du} \\ &= \int{\frac{u^4 + 4u^3 + 6u^2 + 4u + 1}{u}\,du} \\ &= \int{u^3 + 4u^2 + 6u + 4 + \frac{1}{u} \,du} \\ &= \frac{u^4}{4} + \frac{4u^3}{3} + 3u^2 + 4u + \ln{|u|} + C \\ &= \frac{ \left( x - 1 \right)^4 }{4} + \frac{ 4\left( x - 1 \right)^3}{3} + 3\left( x - 1 \right)^2 + 4 \left( x- 1 \right) + \ln{ \left| x - 1 \right| } + C \end{align*}
    Thanks from EliteAndoy and HallsofIvy
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  3. #3
    Junior Member EliteAndoy's Avatar
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    Re: Integrate In anyway possible.

    That also works, but still I'm not certain if my solution is correct. What do you think?
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  4. #4
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    Re: Integrate In anyway possible.

    yes your solution is correct
    the best way to solve it is the one discribed by prove it.

    I suggest you study more the integration ...
    MINOAS
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  5. #5
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    Re: Integrate In anyway possible.

    In such cases i have found that partial fraction approach is always better.
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  6. #6
    Junior Member EliteAndoy's Avatar
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    Re: Integrate In anyway possible.

    Thanks everyone!
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