Integrating Natural Log?

I’m having real problems integrating a natural log. The problem I have been set is (where S = integration sign)

S (x – 2) Ln(3x)

I’m trying to use the integration by parts rule but keep getting the wrong answer and I think it might be to do with the natural log. I have used

f(x) = Ln(3x) f'(x) = 1/x g'(x) = (x – 2) and g(x) = 1/2x^2 – 2x

Using the integration by parts formula, I end up with

Ln(3x) * x^2/2 – 2x + S1/2x – 2x

The answer is wrong….. Can someone help me out here…. Where am I going wrong?

Top 1 Answers

A few days ago

³√carthagebrujah

Favorite Answer

Integration by parts:

∫f(x)g'(x) dx = f(x)g(x) – ∫f'(x)g(x) dx

You defined f(x) and g(x) correctly…but your answer is wrong because:

a. if your answer leaves an integral, you continue integrating.

b. the formula inside the remaining integral should be ∫1/2x – 2 (because 2x * 1/x = 2)

c. integration by parts results in subtraction, not addition of the integral.

Thus:

(Ln(3x) * x²/2 – 2x) – ∫1/2x – 2 dx

(Ln(3x) * x²/2 – 2x) – x²/4 + 2x

Answer (note – your book may distribute out some or all of the parentheses):

(x²/2 – 2x)(ln 3x) – x(x/4 + 2)

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