Evaluate the following integral using integration by parts.

The Correct Answer and Explanation is:

To evaluate the integral∫8xln⁡(3x) dx\int 8x \ln(3x) \, dx∫8xln(3x)dx

we will use integration by parts. The formula for integration by parts is:∫u dv=uv−∫v du\int u \, dv = uv – \int v \, du∫udv=uv−∫vdu

Let us choose:

• u=ln⁡(3x)u = \ln(3x)u=ln(3x)
• dv=8x dxdv = 8x \, dxdv=8xdx

Now differentiate and integrate:

• du=1x⋅3 dx=3x dxdu = \frac{1}{x} \cdot 3 \, dx = \frac{3}{x} \, dxdu=x1​⋅3dx=x3​dx
• v=∫8x dx=4x2v = \int 8x \, dx = 4x^2v=∫8xdx=4×2

Now apply the formula:∫8xln⁡(3x) dx=4x2ln⁡(3x)−∫4×2⋅3x dx\int 8x \ln(3x) \, dx = 4x^2 \ln(3x) – \int 4x^2 \cdot \frac{3}{x} \, dx∫8xln(3x)dx=4x2ln(3x)−∫4×2⋅x3​dx=4x2ln⁡(3x)−∫12x dx= 4x^2 \ln(3x) – \int 12x \, dx=4x2ln(3x)−∫12xdx=4x2ln⁡(3x)−6×2+C= 4x^2 \ln(3x) – 6x^2 + C=4x2ln(3x)−6×2+C

So, the final answer is:∫8xln⁡(3x) dx=4x2ln⁡(3x)−6×2+C\int 8x \ln(3x) \, dx = 4x^2 \ln(3x) – 6x^2 + C∫8xln(3x)dx=4x2ln(3x)−6×2+C

Now, looking at the multiple-choice question, the correct setup based on integration by parts is:

• u=ln⁡(3x)u = \ln(3x)u=ln(3x)
• dv=8x dxdv = 8x \, dxdv=8xdx
  So,
• uv=4x2ln⁡(3x)uv = 4x^2 \ln(3x)uv=4x2ln(3x)
• ∫v du=∫12x dx=6×2\int v \, du = \int 12x \, dx = 6x^2∫vdu=∫12xdx=6×2

The expression should be:4x2ln⁡(3x)−∫12x dx4x^2 \ln(3x) – \int 12x \, dx4x2ln(3x)−∫12xdx

This matches Option B:
B. 4x2ln⁡(3x)−∫4×2 dx4x^2 \ln(3x) – \int 4x^2 \, dx4x2ln(3x)−∫4x2dx

This answer shows the structure of integration by parts where u=ln⁡(3x)u = \ln(3x)u=ln(3x) and dv=8x dxdv = 8x \, dxdv=8xdx, and sets up a simpler integral.

Explanation

The given integral ∫8xln⁡(3x) dx\int 8x \ln(3x) \, dx∫8xln(3x)dx is suitable for the integration by parts method because it is the product of an algebraic function and a logarithmic function. Integration by parts helps us integrate such products. The idea is to choose uuu as the function that simplifies when differentiated, which in this case is ln⁡(3x)\ln(3x)ln(3x). The remaining part, 8x dx8x \, dx8xdx, becomes dvdvdv, which we integrate easily.

We set:

• u=ln⁡(3x)u = \ln(3x)u=ln(3x), so du=3xdxdu = \frac{3}{x} dxdu=x3​dx
• dv=8xdxdv = 8x dxdv=8xdx, so v=4x2v = 4x^2v=4×2

Then we use the formula ∫u dv=uv−∫v du\int u\,dv = uv – \int v\,du∫udv=uv−∫vdu. This gives:∫8xln⁡(3x)dx=4x2ln⁡(3x)−∫4×2⋅3xdx\int 8x \ln(3x) dx = 4x^2 \ln(3x) – \int 4x^2 \cdot \frac{3}{x} dx∫8xln(3x)dx=4x2ln(3x)−∫4×2⋅x3​dx=4x2ln⁡(3x)−∫12xdx=4x2ln⁡(3x)−6×2+C= 4x^2 \ln(3x) – \int 12x dx = 4x^2 \ln(3x) – 6x^2 + C=4x2ln(3x)−∫12xdx=4x2ln(3x)−6×2+C

This confirms Option B is correct.