What is the integral of (3690)/(x^{-99)} ?

The integral of (3690)/(x^{-99)} is (369x^{100})/(10)+C

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integral of (3690)/(x^{-99)}

\int \frac{3690}{x ^{- 99}} d x

\frac{369 x ^{100}}{10} + C

Solution steps

\int \frac{3690}{x ^{- 99}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3690 \cdot \int \frac{1}{x ^{- 99}} d x

Apply the Power Rule

: \frac{x ^{100}}{100}

= 3690 \cdot \frac{x ^{100}}{100}

Simplify

3690 \cdot \frac{x ^{100}}{100} : \frac{369 x ^{100}}{10}

= \frac{369 x ^{100}}{10}

Add a constant to the solution

= \frac{369 x ^{100}}{10} + C

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What is the integral of (3690)/(x^{-99)} ?
The integral of (3690)/(x^{-99)} is (369x^{100})/(10)+C