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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)]
Integral of 1 plus x2?
The integral of 1 + x2 is x + 1/3 x3 + C.
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
Integral of -x 2?
The integral of -x2 is -1/3 x3 .
What is the integral of sqrt1-x2?
-1
What is the double integral of 2x?
The integral of a single term of a polynomial, in the form of AxN is (A/N+1) x (N+1). The first integral of 2x is x2 + C. The second integral of 2x is the first integral of x2 + C, which is 1/3x3 + Cx + C.
What is the integral of x divided by a 7?
First just "factor" out the constant: (1/7)∫ x/7 dx = 1/7∫x dxThe integral of xn is xn+1/ (n+1)In this case, our n=1 so:x(1+1)/(1+1) = x2/2Don't forget about the 1/7 we factored out:(1/7)(1/2)(x2) = x2/14Therefore the answer is (don't forget your C):∫x/7 dx =x2/14 + C
What is the integral of 1 divided by the square root of the quantity 1 plus the square of x with respect to x?
2
What is the integral of 1 divided by the square root of the quantity of the square of x minus 1 with respect to x?
0
The indefinite integral of x2 divided by 4 plus 16?
x3 /12 + 16x + c
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 2-x?
The integral of 2-x = 2x - (1/2)x2 + C.
What is the definite integral of 1 divided by x squared?
For it to be a definite integral, you would need to specify a range. We can however give you the indefinite integral. The easiest way to do this is to think of it not as a fraction, but as a negative exponent: 1/x2 = x-2 It then becomes quite easy to integrate, as we can say in general: ∫(axn) dx = ax(n + 1) / (n + 1) + C In this case then, we have: ∫(x-2) dx = -x-1 + C, or -1/x + C
