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What is the integral of 1 divided by x squared?
The indefinite integral of (1/x^2)*dx is -1/x+C.
Integral of 1 plus x2?
The integral of 1 + x2 is x + 1/3 x3 + C.
What is the integral of 1 divided by xln2x?
0.5
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
Integral of sin 2x divided by 1 plus cosine squared x?
3
How do you integrate the integral of quantity of x plus 2 divided by x squared plus x plus 1?
3
What is the integral of 2 times x divided by the quantity x plus 1 to the third power?
3
What is the integral of 1 divided by x squared?
The indefinite integral of (1/x^2)*dx is -1/x+C.
What is the integral of 1 divided by the quantity 1 plus the square of x with respect to x?
2
Integral of 1 divided by x cubed?
I will assume that this is sopposed to be integrated with respect to x. To make this problem easier, imagine that the integrand is x raised to the negative 3. The integral is 1/(-2x-2) plus some constant c.
Integrate1 plus x divided by x2 plus 1?
(1+x)/(x^2+1) Let x^2+1 =u 2x dx = du x dx = du/2 (1+x) / (x^2+1) = 1/(x^2+1) + x / (x^2+1) Integral of x dx / (x^2+1) = (1/2) integral du / u = 1/2 ln|u| --(1) Integral of 1 / (x^2+1) = arctan(x) --(2) Adding (1) and (2) Integral (1+x)/(x^2+1) = (1/2) ln(x^2+1) + arctan(x) + C
What is the integral of dx divided by x-1 to the power of 2?
-(x-1)-1 or -1/(x-1)
Integral of 1 plus x2?
The integral of 1 + x2 is x + 1/3 x3 + C.
What is the integral of 1 over x squared plus 1?
arctan(x)
What is the integral of 1 divided by xln2x?
0.5
What is the integral of 1 divided by x-2?
3
What is the integral of cosx divided by sinx plus cosx from 0 to 2pi?
The Integral diverges. It has singularities whenever sin(x)+cos(x)=0. Singularities do not necessarily imply that the integral goes to infinity, but that is the case here, since the indefinite integral is x/2 + 1/2 Log[-Cos[x] - Sin[x]]. Obviously this diverges when evaluated at zero and 2pi.
