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If $f(x)=\ln \left(x-\sqrt{1+x^{2}}\right)$, then what is $\int f^{\prime \prime}(x) d x$ equal to?
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Verified Answer
The correct answer is:
$-\frac{1}{\sqrt{1+x^{2}}}+c$
Given that $f(x)=\ln \left(x-\sqrt{1+x^{2}}\right)$
$\int \mathrm{f}^{\prime \prime}(\mathrm{x}) \mathrm{dx}=\mathrm{f}^{\prime}(\mathrm{x})+\mathrm{c}$ where $\mathrm{c}$ is a constant
$=\frac{1}{\left(x-\sqrt{1+x^{2}}\right)} \cdot\left(1-\frac{2 x}{2 \sqrt{1+x^{2}}}\right)+c$
$=\frac{-\left(x-\sqrt{1+x^{2}}\right)}{\left(\sqrt{1+x^{2}}\right)\left(x-\sqrt{1+x^{2}}\right)}+c=-\frac{1}{\sqrt{1+x^{2}}}+c$
$\int \mathrm{f}^{\prime \prime}(\mathrm{x}) \mathrm{dx}=\mathrm{f}^{\prime}(\mathrm{x})+\mathrm{c}$ where $\mathrm{c}$ is a constant
$=\frac{1}{\left(x-\sqrt{1+x^{2}}\right)} \cdot\left(1-\frac{2 x}{2 \sqrt{1+x^{2}}}\right)+c$
$=\frac{-\left(x-\sqrt{1+x^{2}}\right)}{\left(\sqrt{1+x^{2}}\right)\left(x-\sqrt{1+x^{2}}\right)}+c=-\frac{1}{\sqrt{1+x^{2}}}+c$
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