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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
What is the integral of 1 divided by xln2x?
0.5
Integral of sec squared radical x?
integral of radical sinx
What is the integral of cot squared x?
ln(sinx) + 1/3ln(sin3x) + C
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 1 divided by 2x squared?
0.5
What is the integral of 1 divided by the cosine squared of x with respect to x?
∫ 1/cos2(x) dx = tan(x) + C C is the constant of integration.
What is the integral of 1 divided by the sine squared of x with respect to x?
∫ 1/sin2(x) dx = -cot(x) + CC is the constant of integration.
What is the integral of 1 over x squared plus 1?
arctan(x)
What is the integral of -2x divided by the square root of x squared minus 4?
- ln ((x^2)-4)
What is the integral of the square root of x minus 1 squared?
the integral of the square-root of (x-1)2 = x2/2 - x + C
What is x divided by x squared?
x-1
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 equals 1 cos squared x divided by cos squared x?
1. Anything divided by itself always equals 1.
What is the integral of the derivative with respect to x of f divided by the quantity p squared plus q squared f squared with respect to x where f is a function of x and p and q are constants?
∫ f'(x)/(p2 + q2f(x)2) dx = [1/(pq)]arctan(qf(x)/p)
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
