Find the general antiderivative of the function below. G(x) = + C need help with this bad johnny please thanks
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Hello, $\displaystyle \frac{1}{x^3}=x^{-3}$ $\displaystyle \frac{1}{\sqrt{x}}=x^{-1/2}$ The antiderivative for $\displaystyle x^a, \forall a \in \mathbb{R}$ is $\displaystyle \frac{x^{a+1}}{a+1}$
Originally Posted by Moo Hello, $\displaystyle \frac{1}{x^3}=x^{-3}$ $\displaystyle \frac{1}{\sqrt{x}}=x^{-1/2}$ The antiderivative for $\displaystyle x^a, \forall a \in \mathbb{R}$ is $\displaystyle \frac{x^{a+1}}{a+1}$ Except $\displaystyle a=-1$ in which case it is $\displaystyle \ln(x)$
Yep
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