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This is the same as 4x-1, so the answer is just 4 ln x + C.

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Q: How do you integrate 4 over x?
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How do you integrate 2sinxcosx?

Integrate 2sin(x)cos(x)dxLet u = cos(x) and du = -sin(x)dx and pull out the -2:-2[Integral(u*du)]Integrate with respect to u:-2(u2)/2 + CSimplify:-u2 + CReplace u with cos(x):-cos2(x) + C


How do you find the derivative of a function that has been raised to a power greater than 1?

Let us say that f(x)=x^4A derivative is the opposite to an integral.If you were to integrate x^4, the first process is taking the power [which in this case is 4], multiplying it by any value before the x [which is 1], then subtracting 1 from the initial power [4]. This leaves 4x^3. The final step is taking the integral of what is 'inside' the power [which is (x)], and multiplying this to the entire answer, which results in 4x^3 x 1 = 4x^3If you were to derive (x)^4, you would just add 1 to the power [4] to become (x)^5 then put the value of the power as the denominator and the function as a numerator. This leaves [(x^5)/(5)]To assure that the derivative is correct, integrate it. (x^5) would become 5x^4. Since (x^5) is over (5), [(5x^4)/(5)] cancels the 5 on the numerator and denominator, thus leaving the original function of x^4


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Integrate from 0 to 3 for modulus x2-4 dx?

First see if the integrand ie x2-4 is negative anywhere in the range (it might change sign either side of anyplace where the function is zero, so first solve for x2-4=0). If it is, reverse the sign in that part. Here if x<2 it is negative so the modulus gives 4-x2. So integrate 4-x2 from 0 to 2 and then integrate x2-4 from 2 to 3. Add the 2 results together.

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What is the integral of 1 divided over 4x-x2?

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What is the solution to x over 4 minus 4 equals x over 4?

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How do you integrate 2sinxcosx?

Integrate 2sin(x)cos(x)dxLet u = cos(x) and du = -sin(x)dx and pull out the -2:-2[Integral(u*du)]Integrate with respect to u:-2(u2)/2 + CSimplify:-u2 + CReplace u with cos(x):-cos2(x) + C