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The integral of 2x is x^2+c, where c is a constant. If this is a definite integral, meaning that the limits of integration are known, then c=0. If this is an indefinite integral, meaning the limits of integration are unknown, then c should either be left as is or solved for using an initial condition.
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How do you integrate 2x over 2x-2?
Integral of 2x dx /(2x-2) Let 2x=u 2 dx = du dx = (1/2) du Integral of 2x dx /(2x-2) = Integral of (1/2) u du / (u-2) = Integral of 1/2 [ (u-2+2) / (u-2)] dx = Integral of 1/2 [ 1+ 2/(u-2)] dx = u/2 + (1/2) 2 ln(u-2) + C = u/2 + ln(u-2) + C = (2x/2) + ln(2x-2) + C = x + ln(2x-2) + C
Integral of ln2x?
I will denote an integral as \int (LaTeX). We can let u = 2x and du = 2dx, and substitute \int ln (2x) dx = (1/2) \int ln u du. Either using integration by parts or by memorization, this is equal to (1/2) u ln u - u + C = (1/2)(2x ln (2x) - 2x) + C, where C is an arbitrary constant.
Determine the integral 3x minus1 x plus 2dx?
Int[3x-x+2] =Int[2x+2] =x^2 +2x +C
How do you integrate 2 power 2x?
Keep in mind that the integral of ex = ex and 2x = ex / ln 2. You can then rewrite the exponent as u = 2x, convert dx to du, and from there it's pretty straightforward. (I've left off the "+ C" part, because you should just know that.)
How do you do integration math?
Well if you are talking about calculus, integration is the anti-derivative. So as my teacher explained to us, instead of going down, you will go up. For example if you have the F(x) = 2x, the F'(x) = 2. F'(x) is the derivative here, so you will do the anti of a derivative. So with the same F(x) = 2x the integral, is SF(x) = 1/3x^3. The Integral will find you the area under the curve.
