Integrate.

The Correct Answer and Explanation is:

We are asked to evaluate the definite integral:∫15ln⁡(3x) dx\int_1^5 \ln(3x) \, dx∫15​ln(3x)dx

Step 1: Use the logarithmic identity

Recall the identity:ln⁡(3x)=ln⁡3+ln⁡x\ln(3x) = \ln 3 + \ln xln(3x)=ln3+lnx

So we can rewrite the integral as:∫15ln⁡(3x) dx=∫15(ln⁡3+ln⁡x) dx\int_1^5 \ln(3x) \, dx = \int_1^5 (\ln 3 + \ln x) \, dx∫15​ln(3x)dx=∫15​(ln3+lnx)dx

Now break it into two separate integrals:=∫15ln⁡3 dx+∫15ln⁡x dx= \int_1^5 \ln 3 \, dx + \int_1^5 \ln x \, dx=∫15​ln3dx+∫15​lnxdx

Since ln⁡3\ln 3ln3 is a constant:=ln⁡3∫15dx+∫15ln⁡x dx= \ln 3 \int_1^5 dx + \int_1^5 \ln x \, dx=ln3∫15​dx+∫15​lnxdx

The first integral is straightforward:∫15dx=5−1=4\int_1^5 dx = 5 – 1 = 4∫15​dx=5−1=4

So:ln⁡3⋅4=4ln⁡3\ln 3 \cdot 4 = 4 \ln 3ln3⋅4=4ln3

Step 2: Compute ∫15ln⁡x dx\int_1^5 \ln x \, dx∫15​lnxdx

Use integration by parts:

Let
u=ln⁡x⇒du=1xdxu = \ln x \Rightarrow du = \frac{1}{x} dxu=lnx⇒du=x1​dx
dv=dx⇒v=xdv = dx \Rightarrow v = xdv=dx⇒v=x

So:∫ln⁡x dx=xln⁡x−∫x⋅1xdx=xln⁡x−∫1dx=xln⁡x−x+C\int \ln x \, dx = x \ln x – \int x \cdot \frac{1}{x} dx = x \ln x – \int 1 dx = x \ln x – x + C∫lnxdx=xlnx−∫x⋅x1​dx=xlnx−∫1dx=xlnx−x+C

Now evaluate the definite integral:∫15ln⁡x dx=[xln⁡x−x]15\int_1^5 \ln x \, dx = \left[ x \ln x – x \right]_1^5∫15​lnxdx=[xlnx−x]15​

Evaluate at the limits:

At x=5x = 5x=5:
5ln⁡5−55 \ln 5 – 55ln5−5

At x=1x = 1x=1:
1ln⁡1−1=0−1=−11 \ln 1 – 1 = 0 – 1 = -11ln1−1=0−1=−1

So:∫15ln⁡x dx=(5ln⁡5−5)−(−1)=5ln⁡5−4\int_1^5 \ln x \, dx = (5 \ln 5 – 5) – (-1) = 5 \ln 5 – 4∫15​lnxdx=(5ln5−5)−(−1)=5ln5−4

Step 3: Combine everything

∫15ln⁡(3x) dx=4ln⁡3+(5ln⁡5−4)\int_1^5 \ln(3x) \, dx = 4 \ln 3 + (5 \ln 5 – 4)∫15​ln(3x)dx=4ln3+(5ln5−4)=4ln⁡3+5ln⁡5−4= 4 \ln 3 + 5 \ln 5 – 4=4ln3+5ln5−4

Step 4: Approximate

Use calculator values:

• ln⁡3≈1.0986\ln 3 \approx 1.0986ln3≈1.0986
• ln⁡5≈1.6094\ln 5 \approx 1.6094ln5≈1.6094

So:4ln⁡3≈4⋅1.0986=4.39444 \ln 3 \approx 4 \cdot 1.0986 = 4.39444ln3≈4⋅1.0986=4.39445ln⁡5≈5⋅1.6094=8.04705 \ln 5 \approx 5 \cdot 1.6094 = 8.04705ln5≈5⋅1.6094=8.04704.3944+8.0470−4=8.44144.3944 + 8.0470 – 4 = 8.44144.3944+8.0470−4=8.4414

Final Answer:

∫15ln⁡(3x) dx≈8.4414\int_1^5 \ln(3x) \, dx \approx \boxed{8.4414}∫15​ln(3x)dx≈8.4414​