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How do you integrate #ln(3x)#?

Answer 1

Hello,

Answer : #x ln(3x) - x + cte#.

Remark that #d/dx (X ln(X) - X) = 1.ln(X) + X/X - 1 = ln(X)#, therefore #int ln(X) dX = X ln(X) - X + cte#

Now, it's easy because, #ln(3x) = ln(3) + ln(x)#, therefore, #int ln(3x) dx = ln(3)x + x ln(x) - x + cte = x ln(3x) - x + cte#

PS. There is a classical trick to find #int ln(x) dx# if you don't have any idea to test #x ln(x) -x# : an integration by parts. The general formula is #int u'v = uv - int uv'# (because #(uv)' = u'v + uv'#). Now, write #int 1 \cdot ln(x) dx = \int u' v#, with #u=x# and #v=ln(x)#. You get #int ln(x) dx = x ln(x) - \int x/x dx = x ln(x) -x + cte#

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Answer 2

Evelyn Agard

To integrate ln(3x), you can use integration by parts. Let u = ln(3x) and dv = dx. Then, du = (1/x) dx and v = x. Applying the integration by parts formula, the integral becomes uv - ∫v du. Plugging in the values, you get x ln(3x) - ∫x (1/x) dx. Simplifying the integral, you have x ln(3x) - ∫1 dx. Integrating 1 dx gives x ln(3x) - x + C, where C is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.