Assertion (A): $\int \frac{dx}{x} = \log|x| + C$ on every interval that does not contain 0, negative $x$ included.
Reason (R): $\frac{d}{dx}\log|x| = \frac{1}{x}$ for every $x \ne 0$ , negative $x$ included.

Both A and R are true, and R is the correct explanation of A.Correct answerYour answerBoth A and R are true, but R is not the correct explanation of A.Correct answerYour answerA is true, but R is false.Correct answerYour answerA is false, but R is true.Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer A

Correct answer: A

For $x < 0$ , $\log|x| = \log(-x)$ , whose derivative is $\frac{-1}{-x} = \frac{1}{x}$ .
So R is true, and it is exactly why A holds for negative $x$ as well: A is true and R explains it.

Concept: The absolute value in log|x|
Typical timeTypical: 90 s
Your timeYours: —

The rule this testsIntegration as the reverse of differentiation

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Ten questions from Indefinite Integrals, at your own pace.

Assertion (A): ∫dxx=log⁡∣x∣+C\int \frac{dx}{x} = \log|x| + C∫xdx​=log∣x∣+C on every interval that does not contain 0, negative xxx included.Reason (R): ddxlog⁡∣x∣=1x\frac{d}{dx}\log|x| = \frac{1}{x}dxd​log∣x∣=x1​ for every x≠0x \ne 0x=0, negative xxx included.

Explanation

Assertion (A): $\int \frac{dx}{x} = \log|x| + C$ on every interval that does not contain 0, negative $x$ included.
Reason (R): $\frac{d}{dx}\log|x| = \frac{1}{x}$ for every $x \ne 0$ , negative $x$ included.