Evaluate ₁0} x \, dx

Correct Answer is : log⁡10e(xlog⁡x−x)+c₁0} e (x x - x) + clog10e(xlogx−x)+c

Correct Answer is : log⁡10e(xlog⁡x−x)+c₁0} e (x x - x) + clog10e(xlogx−x)+c

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Integral Calculus

Question : 35 of 73

Marks: +1, -0

Evaluate

\int\limits \log_{10} x \, dx

Solution:

Concept:

Rewrite

\log_{10} x

\log_{10} e \cdot \ln x

and integrate

\ln x

Explanation:

We have

\log_{10} x = \frac{\ln x}{\ln 10} = (\log_{10} e) \ln x

Thus

\int \log_{10} x \, dx = \log_{10} e \int \ln x \, dx

We know

\int \ln x \, dx = x \ln x - x + c

Therefore,

\int \log_{10} x \, dx = \log_{10} e \, (x \ln x - x) + c

Here

\ln x

is the natural logarithm (often written as

\log x

in option B).

Answer:

Option B:

\log_{10} e (x \log x - x) + c

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