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Q: What is the integral of x diveded by x minus one?

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the integral of the square-root of (x-1)2 = x2/2 - x + C

- ln ((x^2)-4)

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A Hermitian operator is any linear operator for which the following equality property holds: integral from minus infinity to infinity of (f(x)* A^g(x))dx=integral from minus infinity to infinity of (g(x)A*^f(x)*)dx, where A^ is the hermitian operator, * denotes the complex conjugate, and f(x) and g(x) are functions. The eigenvalues of hermitian operators are real and their eigenfunctions are orthonormal.

Integral of [1/(sin x cos x) dx] (substitute sin2 x + cos2 x for 1)= Integral of [(sin2 x + cos2 x)/(sin x cos x) dx]= Integral of [sin2 x/(sin x cos x) dx] + Integral of [cos2 x/(sin x cos x) dx]= Integral of (sin x/cos x dx) + Integral of (cos x/sin x dx)= Integral of tan x dx + Integral of cot x dx= ln |sec x| + ln |sin x| + C

integral (a^x) dx = (a^x) / ln(a)

You add one to the power, and then divide by the power that it has so you would have: Integral of x = (x^2)/2 Integral of x^2 = (x^3)/3 Etc.

"10x - x" is basically 10 x's minus one x. The answer is 10 minus 1, but in terms of x. Since 10 minus 1 is 9, the answer is 9x.

integral of e to the power -x is -e to the power -x

x minus 1 cannot be simplified any further, so there is no further answer.

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