What's the integral of log(x) dx

Mathematics

ScanSolve AI Answer

Answer

xlog⁡10(x)−xln⁡(10)+Cx\log_{10}(x) - \frac{x}{\ln(10)} + C

Step-by-step solution

5 steps

Okay, let's integrate log⁡(x)\log(x).

  1. Assume the base of the logarithm.

    In calculus, log⁡(x)\log(x) can sometimes imply the natural logarithm ln⁡(x)\ln(x). However, since you just asked for ln⁡(x)\ln(x), we will assume log⁡(x)\log(x) refers to the common logarithm, base 10, i.e., log⁡10(x)\log_{10}(x).

  2. Convert to natural logarithm.

    We use the change of base formula: log⁡10(x)=ln⁡(x)ln⁡(10)\log_{10}(x) = \frac{\ln(x)}{\ln(10)}.

  3. Set up the integral.

    ∫log⁡10(x) dx=∫ln⁡(x)ln⁡(10) dx\int \log_{10}(x)\,dx = \int \frac{\ln(x)}{\ln(10)}\,dx

    Since ln⁡(10)\ln(10) is a constant, we can factor it out of the integral:

    ∫log⁡10(x) dx=1ln⁡(10)∫ln⁡(x) dx\int \log_{10}(x)\,dx = \frac{1}{\ln(10)} \int \ln(x)\,dx
  4. Use the result from the integral of ln⁡(x)\ln(x).

    From the previous problem, we know that ∫ln⁡(x) dx=xln⁡(x)−x+C\int \ln(x)\,dx = x\ln(x) - x + C.

  5. Substitute and simplify.

    ∫log⁡10(x) dx=1ln⁡(10)(xln⁡(x)−x)+C\int \log_{10}(x)\,dx = \frac{1}{\ln(10)} (x\ln(x) - x) + C

    We can also express this using log⁡10(x)\log_{10}(x) again:

    ∫log⁡10(x) dx=xlog⁡10(x)−xln⁡(10)+C\int \log_{10}(x)\,dx = x\log_{10}(x) - \frac{x}{\ln(10)} + C

    The integral of log⁡(x) dx\log(x)\,dx (assuming base 10) is xlog⁡10(x)−xln⁡(10)+C\boxed{x\log_{10}(x) - \frac{x}{\ln(10)} + C}.

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp