The Correct Answer and Explanation is:

We are asked to evaluate the integral∫3xln⁡(3x) dx\int 3x \ln(3x) \, dx∫3xln(3x)dx

using integration by parts.


Step 1: Use substitution to simplify

Letu=3x⇒du=3dx⇒dx=13duu = 3x \Rightarrow du = 3dx \Rightarrow dx = \frac{1}{3} duu=3x⇒du=3dx⇒dx=31​du

Substitute into the integral:∫3xln⁡(3x) dx=∫uln⁡(u)⋅13 du=13∫uln⁡(u) du\int 3x \ln(3x) \, dx = \int u \ln(u) \cdot \frac{1}{3} \, du = \frac{1}{3} \int u \ln(u) \, du∫3xln(3x)dx=∫uln(u)⋅31​du=31​∫uln(u)du


Step 2: Apply integration by parts

Use the integration by parts formula:∫u dv=uv−∫v du\int u \, dv = uv – \int v \, du∫udv=uv−∫vdu

Let:

  • a=ln⁡(u)⇒da=1udua = \ln(u) \Rightarrow da = \frac{1}{u} dua=ln(u)⇒da=u1​du
  • db=u du⇒b=u22db = u \, du \Rightarrow b = \frac{u^2}{2}db=udu⇒b=2u2​

Now apply the formula:∫uln⁡(u) du=u22ln⁡(u)−∫u22⋅1udu\int u \ln(u) \, du = \frac{u^2}{2} \ln(u) – \int \frac{u^2}{2} \cdot \frac{1}{u} du∫uln(u)du=2u2​ln(u)−∫2u2​⋅u1​du=u22ln⁡(u)−∫u2du=u22ln⁡(u)−14u2= \frac{u^2}{2} \ln(u) – \int \frac{u}{2} du = \frac{u^2}{2} \ln(u) – \frac{1}{4} u^2=2u2​ln(u)−∫2u​du=2u2​ln(u)−41​u2


Step 3: Multiply by the constant from substitution

13(u22ln⁡(u)−14u2)=u26ln⁡(u)−u212\frac{1}{3} \left( \frac{u^2}{2} \ln(u) – \frac{1}{4} u^2 \right) = \frac{u^2}{6} \ln(u) – \frac{u^2}{12}31​(2u2​ln(u)−41​u2)=6u2​ln(u)−12u2​

Recall that u=3xu = 3xu=3x. So:u2=(3x)2=9x2u^2 = (3x)^2 = 9x^2u2=(3x)2=9×2∫3xln⁡(3x) dx=9x26ln⁡(3x)−9×212+C\int 3x \ln(3x) \, dx = \frac{9x^2}{6} \ln(3x) – \frac{9x^2}{12} + C∫3xln(3x)dx=69×2​ln(3x)−129×2​+C

Simplify:=32x2ln⁡(3x)−34×2+C= \frac{3}{2} x^2 \ln(3x) – \frac{3}{4} x^2 + C=23​x2ln(3x)−43​x2+C


Final Answer:

∫3xln⁡(3x) dx=32x2ln⁡(3x)−34×2+C\boxed{\int 3x \ln(3x) \, dx = \frac{3}{2} x^2 \ln(3x) – \frac{3}{4} x^2 + C}∫3xln(3x)dx=23​x2ln(3x)−43​x2+C​

This method uses substitution followed by integration by parts. The key idea is to simplify the integral before applying the formula. This helps in correctly identifying the parts and avoiding complex expressions. The natural log function integrates well using this approach because its derivative is simpler, making it a good choice for the uuu part.

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