Q. \[ \int \frac{1}{x}\,dx \]

Q: What is the integral of \frac{1}{x} ?

The antiderivative is \int \frac{1}{x}\,dx = \ln|x| + C , valid for x \ne 0.

Q: Why do we use \ln|x| instead of \ln x ?

\ln x is only defined for x>0. Using \ln|x| covers x<0 too, since \frac{d}{dx}\ln|x|=\frac{1}{x} for x\ne0.

Q: What happens if x=0 in \int \frac{1}{x}\,dx ?

x=0 is undefined because \frac{1}{x} diverges. The antiderivative \ln|x|+C is defined on intervals that do not cross 0.

Q: What is \int \frac{1}{x^2}\,dx and how is it different from \int \frac{1}{x}\,dx ?

\int \frac{1}{x^2}\,dx = -\frac{1}{x}+C . Only \frac{1}{x} leads to a logarithm.

Q: How do you compute a definite integral like \int_a^b \frac{1}{x}\,dx ?

For a,b\ne0 and not crossing 0, \int_a^b \frac{1}{x}\,dx = \ln|b|-\ln|a| = \ln\left|\frac{b}{a}\right| .

Q: What about \int_{{-1}}^{1}\frac{1}{x}\,dx ?

The improper integral diverges because the integrand has a non-removable singularity at x=0. The antiderivative \ln|x| goes to -\infty near 0.

Q: Is \int \frac{1}{x}\,dx = \log_{10}|x| + C also correct ?

Yes, up to a constant factor: \ln|x| = \log_{10}|x|\cdot\ln(10) . Any base works, but the derivative must match \frac{1}{x} .

Answer

We evaluate the indefinite integral:

\[
\int \frac{1}{x}\,dx
\]

Since \(\frac{1}{x} = x^{-1}\),

\[
\int x^{-1}\,dx = \ln|x| + C
\]

Final result:

\[
\ln|x| + C
\]

Detailed Explanation

We want to find the indefinite integral

\[
\int \frac{1}{x}\, dx.
\]

Step 1: Identify the integrand in a standard form.

We recognize that

\[
\frac{1}{x} = x^{-1}.
\]

So the integral becomes

\[
\int x^{-1}\, dx.
\]

Step 2: Use the power rule for integrals (with a special case).

The power rule says: for \(n \neq -1\),

\[
\int x^{n}\, dx = \frac{x^{n+1}}{n+1} + C.
\]

But here we have \(n = -1\). That means the usual formula would require dividing by \(n+1 = 0\), which is not allowed. So we use the special logarithm fact:

\[
\int \frac{1}{x}\, dx = \ln|x| + C.
\]

Step 3: State the final answer.

\[
\int \frac{1}{x}\, dx = \ln|x| + C.
\]

Final result:

\[
\boxed{\ln|x| + C}
\]

See full solution

Graph

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Calculus FAQ

What is the integral of \( \frac{1}{x} \) ?

The antiderivative is \( \int \frac{1}{x}\,dx = \ln|x| + C \), valid for \(x \ne 0\).

Why do we use \( \ln|x| \) instead of \( \ln x \) ?

\( \ln x \) is only defined for \(x>0\). Using \( \ln|x| \) covers \(x<0\) too, since \( \frac{d}{dx}\ln|x|=\frac{1}{x} \) for \(x\ne0\).

What happens if \(x=0\) in \( \int \frac{1}{x}\,dx \) ?

\(x=0\) is undefined because \( \frac{1}{x} \) diverges. The antiderivative \( \ln|x|+C \) is defined on intervals that do not cross \(0\).

What is \( \int \frac{1}{x^2}\,dx \) and how is it different from \( \int \frac{1}{x}\,dx \) ?

\( \int \frac{1}{x^2}\,dx = -\frac{1}{x}+C \). Only \( \frac{1}{x} \) leads to a logarithm.

How do you compute a definite integral like \( \int_a^b \frac{1}{x}\,dx \) ?

For \(a,b\ne0\) and not crossing \(0\), \( \int_a^b \frac{1}{x}\,dx = \ln|b|-\ln|a| = \ln\left|\frac{b}{a}\right| \).

What about \( \int_{{-1}}^{1}\frac{1}{x}\,dx \) ?

The improper integral diverges because the integrand has a non-removable singularity at \(x=0\). The antiderivative \( \ln|x| \) goes to \(-\infty\) near \(0\).

Is \( \int \frac{1}{x}\,dx = \log_{10}|x| + C \) also correct ?

Yes, up to a constant factor: \( \ln|x| = \log_{10}|x|\cdot\ln(10) \). Any base works, but the derivative must match \( \frac{1}{x} \).

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