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Hello, Can anyone show me how this step is done? integral of (6-4ln(3-x))^2 dx The next step says to apply linearity: 4 integral sign (2ln(3-x) -3)^2 dx I don't understand how to apply linearity and could not find anything on the internet that explained it. Thanks. |
June 11, 2021 |
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For the first one, I would first expand the square as a polynomial, giving 3 integrals, one of a contant, one of ln(u) and one of ln 2 (u) with u = 3 - x and so du = -dx. Then each of the three integrals is relatively standard, well, more or less, with integral( ln(x) dx ) = x(ln(x) -1) + C and integral( ln 2 (x) dx ) = x ( ln 2 (x) - 2 ln(x) + 2 ) + C I don't understand the second case, since, in the reals, a square is always positive, and thus, the sign is always +1 ( or locally 0). So, as I see it, it is about 4 integral(dx) == 4x + C. (Sure, initially, x has to be smaller than 3, in the Reals, because the log of a negative argument is out of domain). |
June 12, 2021 |
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Thank you for your help. I appreciate it. |
June 13, 2021 |
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