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Explain why
$$\int\frac{f'(x)}{f(x)}\,\mathrm dx = \ln|f(x)|+C.$$
Thanks!

Stefan Hansen
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Ofir Attia
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2 Answers2

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By the Chain rule $$ \frac{\mathrm d}{\mathrm d x}\big(\ln|f(x)|+C\big)=\frac{1}{f(x)}\cdot\frac{\mathrm d}{\mathrm dx}f(x)=\frac{f'(x)}{f(x)} $$

Stefan Hansen
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  • I want to calculate $∫tg(x)dx$ with this rule. so it will be $∫\frac{sin(X)}{(cos(X))}$ than I just need to add minus before the ln? – Ofir Attia Apr 29 '13 at 07:59
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    Yes, $$\int \frac{\sin x}{\cos x},\mathrm dx=-\int\frac{f'(x)}{f(x)},\mathrm dx$$ with $f(x)=\cos x$. – Stefan Hansen Apr 29 '13 at 08:09
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Differentiate RHS: you get $\frac{d}{dx}(\log |f(x)|+C)=\frac{d}{dx}(\log |f(x)|)=\frac{1}{|f(x)|}\cdot\frac{d}{dx}|f(x)|=\frac{f'(x)}{|f(x)|}\cdot Sgn(f(x))=\frac{f'(x)}{f(x)}$, where $Sgn(f)=1$ where $f\geq 0$, $Sgn(f)=-1$ where $f<0$

Federica Maggioni
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