The indefinite integral of $\frac{1}{x}$ with respect to x is: $\ln (1/x) + C, x > 0$ $e^x$ $\ln (x) + C$ $\ln (-x) + C$ if $x < 0$
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The indefinite integral of $\frac{1}{x}$ with respect to $x$ is $\ln|x| + C$. Show more…
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Assuming that $x<0$, use differentiation to justify the formula $\int \frac{1}{x} d x=\ln |x|+C$ [Hint: Use the chain rule after noting that $\ln |x|=\ln (-x)$ for $x<0 .]$
Integration
Antiderivatives and Indefinite Integrals
$\left.\int e^{x+\frac{1}{x}}-\frac{e^{x+\frac{1}{x}}}{x^{2}}\right) d x$ is (a) $e^{x+\frac{1}{x}}+C$ (b) $\frac{\mathrm{e}^{\mathrm{x}+\frac{1}{\mathrm{x}}}}{\mathrm{x}}+\mathrm{C}$ (c) $-e^{x+\frac{1}{x}}+C$ (d) $\frac{-e^{x+\frac{1}{x}}}{x}+C$
28. This problem investigates why Theorem 5.1.2 states that ∫ 1/x dx = ln |x| + C. (a) What is the domain of y = ln x? (b) Find d/dx ( ln x ). (c) What is the domain of y = ln(−x)? (d) Find d/dx ( ln(−x) ). (e) You should find that 1/x has two types of antiderivatives, depending on whether x > 0 or x < 0. In one expression, give a formula for ∫ 1/x dx that takes these different domains into account, and explain your answer.
Adi S.
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