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How do you evaluate the integral of 2udu divided by 1 plus u squared?
for the integral of (2x)dx/(1+x2 ) Take (1+x2 ) as your 'u' substitution. find du, du= 2x dx use u substitution to write new integral, integral of du/u the integral of du/u= ln abs(u) + C therefore, your original problem becomes an answer with ln ln abs (1+x2) + C *abs refers to absolute value of the parentheses
What is the integral of sqrt1-x2?
-1
How do you bisect the area of definate integral 1 to 6 of 1 divided by x sqared?
First, find the area under the curve y = 1/x2, with boundary lines x = 1 and x = 6, by calculating the integral of 1/x2 with lower limit 1, and upper limit 6. Then divide it by 2. (6)integral(1) of (1/x2) dx = (6)integral(1) of (x--2) dx = -x-1|(6),(1) = -1/x|(6)(1) = -1/6 +1 = 5/6. Thus, the half of the area under the curve is 5/12.
What is the integral of 1 divided by the quantity 1 plus the square of x with respect to x?
2
What is the integral of rootsquare 1 plus x2 over x without using partial fraction?
Int sqrt(1+x2)/x = sqrt(1+x2) + LN [(sqrt(1+x2) - x -1) / (sqrt(1+x2) - x +1)]
