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Given that d/dx ((ln2x)/3x)= (1-ln2x)/(3x^2), find the integral of ln(2x)/(x^2)?
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d/dx ((ln2x)/3x)= (1-ln2x)/(3x^2)
This means:
∫ (1-ln2x)/(3x^2) dx = ln(2x) / 3x
∫ dx/ (3x^2) - ∫ ln(2x) / (3x^2) dx = ln(2x) / 3x
(1/3)∫ dx/ x^2 -(1/3) ∫ ln(2x) / x^2 dx = ln(2x) / 3x
multiply everything by 3
∫ dx/ x^2 - ∫ ln(2x) / x^2 dx = ln(2x) / x
∫ ln(2x) / x^2 dx = ∫ dx/ x^2 - ln (2x) / x
∫ ln(2x) / x^2 dx = -1/ x - ln (2x) / x
∫ ln(2x) / x^2 dx = -(1+ln (2x) ) / x + C
ln(2x)/(x^2) = 1/x^2 - d/dx((ln2x)/x))
and integrate both sides