Mahdi Z. answered • 11/21/19
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∫[sinx + 1/x - 4x-3]dx = -cosx + ln lxl + 2/x2 + Const
Key J.
asked • 11/20/19Determine the Integral
∫((x4sinx+x3-4x)/(x4))
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Mark M. answered • 11/20/19
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Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
Divide each term of the numerator by x4. We obtain:
∫[sinx + 1/x - 4x-3]dx = -cosx + ln lxl + 2/x2 + C
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