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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.
What is the integral of x divided by x plus 1?
x/(x+1) = 1 - 1/(x + 1), so the antiderivative (or indefinite integral) is x + ln |x + 1| + 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
Integral of 1 plus x2?
The integral of 1 + x2 is x + 1/3 x3 + C.
What are the different examples of integral exponents?
... -3, -2, -1, 0, 1, 2, 3, ...In summary, any integer that you use as an exponent is an "integral exponent".... -3, -2, -1, 0, 1, 2, 3, ...In summary, any integer that you use as an exponent is an "integral exponent".... -3, -2, -1, 0, 1, 2, 3, ...In summary, any integer that you use as an exponent is an "integral exponent".... -3, -2, -1, 0, 1, 2, 3, ...In summary, any integer that you use as an exponent is an "integral exponent".
