Given dy/dx=squareroot(e^x -1), what is the function y?

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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To determine the primitive function y, we'll have to use the indefinite integral:

dy = sqrt(e^x - 1)dx

We'll integrate both sides:

Int dy = Int sqrt(e^x - 1)dx

We'll use substitution technique and we'll replace sqrt(e^x - 1) by t:

sqrt(e^x - 1) = t => e^x = t^2 + 1

We'll differentiate both sides, with respect to x:

e^x dx/2sqrt(e^x - 1) = dt

e^x dx = 2sqrt(e^x - 1)dt

dx = 2sqrt(e^x - 1)dt/e^x

dx = 2tdt/(t^2 + 1)

We'll re-write the integral:

Int sqrt(e^x - 1)dx = Int 2t^2 dt /(t^2 + 1)

Int 2t^2 dt /(t^2 + 1) = 2Int (t^2 + 1 - 1) dt /(t^2 + 1)

Int 2t^2 dt /(t^2 + 1) = 2Int dt - 2Int dt /(t^2 + 1)

Int 2t^2 dt /(t^2 + 1) = 2t - 2arctan t + C

Int sqrt(e^x - 1)dx = 2sqrt(e^x - 1) - 2arctansqrt(e^x - 1) + C

The requested primitive function y is:  y=Int sqrt(e^x - 1)dx = 2sqrt(e^x - 1) - 2arctansqrt(e^x - 1) + C.

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