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Math Help - Integral 2x^2 ln(x^2)

  1. #1
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    Integral 2x^2 ln(x^2)

    I got problem to integral ln (x^2)...

    As the primitive of In (x) = x(ln (x) -1)
    I was wondering... Since ln (x^2) = 2 ln (x), could it work in this way to solve an integral???

    The answer is: e^2 + 1

    Thx in adv for your explanation
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  2. #2
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    Quote Originally Posted by dianjoey View Post
    I got problem to integral ln (x^2)...

    As the primitive of In (x) = x(ln (x) -1)
    I was wondering... Since ln (x^2) = 2 ln (x), could it work in this way to solve an integral???

    The answer is: e^2 + 1

    Thx in adv for your explanation
    2x\ln(x^2) = 2x \cdot 2\ln{x} = 4x \ln{x}

    4\int_1^e x\ln{x} \, dx

    integration by parts ...

    u = \ln{x}

    dv = x \, dx

    proceed ...
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  3. #3
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    Hi Skeeter,

    But the answer is e^2+1...
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  4. #4
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    Quote Originally Posted by dianjoey View Post
    Hi Skeeter,

    But the answer is e^2+1...
    I agree ... you need to finish it.
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  5. #5
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    Skeeter,

    I am just re-doing my old exercises, I have all steps to get the answer, but I just don't understand how to integral ln (x^2)
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  6. #6
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    I will show...

    I provide the answer I want those who is willing to explain can check.
    Attached Thumbnails Attached Thumbnails Integral 2x^2 ln(x^2)-untitled.jpg  
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  7. #7
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    Quote Originally Posted by skeeter View Post
    2x\ln(x^2) = 2x \cdot 2\ln{x} = 4x \ln{x}

    4\int_1^e x\ln{x} \, dx

    integration by parts ...

    u = \ln{x}

    dv = x \, dx

    proceed ...
    Now we have 4\cdot\frac{x^2}{2}\ln(x)-4\int\frac{x^2}{2}\cdot\frac{1}{x}\,dx = 2x^2\ln(x)-4\int\frac{x}{2}\,dx

    Can you finish?

    Spoiler:
    =2x^2\ln(x)-x^2 = \left[x^2(2\ln(x)-1)\right]_1^e = e^2(2\ln(e)-1) - 1^2(2\ln(1)-1) = e^2(2-1) - 1(0-1) = e^2+1, as desired.
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  8. #8
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    Thx Redsoxfan325, very clear step and you way of solving the integral is easier to understand.
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  9. #9
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    Quote Originally Posted by dianjoey View Post
    [snip]
    Since ln (x^2) = 2 ln (x),

    [snip]
    As small point ..... ln(x^2) = 2 ln |x|.

    But since you're working with x > 0 there is no difference between ln x and ln |x| in this question.
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