If the primitive of $$1ex$$−$$12$$ $$isf(x)$$−log⁡|$$g(x)$$|$$+C$$ then

Correct option is 3dom⁡f=R~{0}

Questions

If the primitive of $\frac{1}{{(e^{x} - 1)}^{2}}$ is
$f (x) - \log | g (x) | + C$ then

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dom⁡f=R

g(x)=1−ex

dom⁡f=R~{0}

f(x)=1−e−x

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detailed solution

Correct option is C

∫1ex−12dx=∫exdxexex−12=∫dtt(t−1)2t=ex=∫1t−11t−1−1tdt=∫1(t−1)2−1t−1+1tdt=11−t−log⁡|t−1|+log⁡|t|+C=1−ex−1−log⁡1−e−x+CHence f(x)=1−ex−1 and g(x)=1−e−xThe domain of f=R~{0}

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If the primitive of 1ex−12 isf(x)−log⁡|g(x)|+C then

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

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If the primitive of $\frac{1}{{(e^{x} - 1)}^{2}}$ is
$f (x) - \log | g (x) | + C$ then