My question is about number 17. It shows the question first then next step it goes to $\frac{1}{x^2}$. Can't I just leave that as is instead of making it $x^{-2}$, take anti derivative which is $\ln (x^2)$?
Integrals homework help question
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$\begingroup$
calculus
integration
definite-integrals
indefinite-integrals
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3No, the antiderivative of $\frac{1}{x^2}$ is not $\ln x^2$. – 2017-01-15
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1Note that the antiderivative is only $\ln|x|+C$ for functions in the form $\frac{1}{x^p}$ when $p=1$. – 2017-01-15
2 Answers
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we have $$-\int x^{-2}dx$$ from $$\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$$ if $$n\ne -1$$ we get $$x^{-1}+C$$
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The antiderivative of $1/x^2$ is not $\ln x^2$. In general, the antiderivative of $1/\mbox{crud}$ is not $\ln \mbox{crud}.$ Only the antiderivative of $1/x$ is $\ln x$.