Q. \(\displaystyle \int x^{-1}\,dx\)

Q: What is \\int x^{-1} \\, dx ?

\displaystyle \\int x^{-1} \\, dx = \\ln|x| + C.

Q: Why does the answer involve absolute value |x| ?

\ln(x) is only defined for x>0. Using \\ln|x| ensures the derivative is 1/x for both x>0 and x<0, excluding x=0.

Q: What is \\int \\frac{1}{x} \\, dx ?

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

Q: What happens if I differentiate the result \\ln|x| + C ?

\displaystyle \\frac{d}{dx}(\\ln|x|)=\\frac{1}{x}. The constant C differentiates to 0.

Q: Is there a simpler method to integrate x^{-1} ?

Use the power rule: for n\\neq -1, \\int x^n dx=\\frac{x^{n+1}}{n+1}+C. Since n=-1, it becomes \\ln|x|+C.

Q: What is the definite integral \\int_{a}^{b} \\frac{1}{x} \\, dx ?

For a,b\\neq 0 and same sign: \displaystyle \\int_{a}^{b} \\frac{1}{x} dx = \\ln|b| - \\ln|a| = \\ln\\left|\\frac{b}{a}\\right|.

Answer

We need to find

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

Since \(x^{-1}=\frac{1}{x}\), we use the rule \(\int \frac{1}{x}\,dx=\ln|x|+C\).

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

Detailed Explanation

We want to compute the integral

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

Step 1: Rewrite the power in a more familiar form.

The exponent \( -1 \) means we have the reciprocal of \(x\):

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

So the integral becomes

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

Step 2: Use the standard logarithm integral rule.

A key fact from calculus is that

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

Step 3: State the final answer.

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

See full solution

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Calculus FAQ

What is \(\\int x^{-1} \\, dx\) ?

\(\displaystyle \\int x^{-1} \\, dx = \\ln|x| + C\).

Why does the answer involve absolute value \( |x| \) ?

\(\ln(x)\) is only defined for \(x>0\). Using \(\\ln|x|\) ensures the derivative is \(1/x\) for both \(x>0\) and \(x<0\), excluding \(x=0\).

What is \(\\int \\frac{1}{x} \\, dx\) ?

\(\displaystyle \\int \\frac{1}{x} \\, dx = \\ln|x| + C\).

What happens if I differentiate the result \(\\ln|x| + C\) ?

\(\displaystyle \\frac{d}{dx}(\\ln|x|)=\\frac{1}{x}\). The constant \(C\) differentiates to \(0\).

Is there a simpler method to integrate \(x^{-1}\) ?

Use the power rule: for \(n\\neq -1\), \(\\int x^n dx=\\frac{x^{n+1}}{n+1}+C\). Since \(n=-1\), it becomes \(\\ln|x|+C\).

What is the definite integral \(\\int_{a}^{b} \\frac{1}{x} \\, dx\) ?

For \(a,b\\neq 0\) and same sign: \(\displaystyle \\int_{a}^{b} \\frac{1}{x} dx = \\ln|b| - \\ln|a| = \\ln\\left|\\frac{b}{a}\\right|\).

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