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In order to work out this problem, we need to learn how to apply the integration method correctly.

The given expression is ∫ 2xln(2x) dx.

Instead of working out with 2x's, we let u = 2x. Then, du = 2 dx or du/2 = dx. This method is both valid and easy to avoid working out with too much expressions. You should get:

∫ uln(u) (du/2)

= ½ ∫ uln(u) du

Use integration by parts, which states that:

∫ f(dg) = fg - ∫ g(df)

We let:

  • f = ln(u). Then, df = 1/u du
  • dg = u du. Then, g = ∫ u du = ½u²

Using these substitutions, we now have:

½(½u²ln(u) - ½∫ u du)

= ¼(u²ln(u) - ∫ u du)

Finally, by integration, we obtain:

¼ * (u²ln(u) - ½u²) + c

= 1/8 * (2u²ln(u) - u²) + c

= 1/8 * (2(2x)²ln(u) - (2x)²) + c

= 1/8 * (2x)² * (2ln(u) - 1) + c

= ½ * x² * (2ln(2x - 1)) + c

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Q: What is the indefinite integral of 2x ln2x dx?
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