Question #935b8

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Question #935b8

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Answer 1 · 2016-12-28T14:36:10

You don't. By inspection and checking by differentiation, the integral is $\sqrt{x^{2} + 9} + C$ $sqrt(x^2+9)+C$

Explanation:

The key point is that the numerator is, give or take a constant multiplier, the derivative of the expression under the square root in the denominator. So, thinking of the the square root as ${(x^{2} + 9)}^{- \frac{1}{2}}$ $(x^2+9)^(-1/2)$ , you should immediate think of "adding 1' to the power, giving ${(x^{2} + 9)}^{+ \frac{1}{2}}$ $(x^2+9)^(+1/2)$ . Then, mentally differentiating this last expression, you find, hey presto, that by the chain rule the $x^{2}$ $x^2$ becomes $2 x$ $2x$ which nicely cancels the $2$ $2$ and introduces the $x$ $x$ .

Perhaps you didn't mean to include the $x$ $x$ in the numerator?

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Question #935b8

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