∫ (1/X2 +4X+13) dX
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Originally Posted by mindblowing ∫ (1/X2 +4X+13) dX What do you mean? 1) $\displaystyle \frac{1}{x^2+4x+13}$ 2) $\displaystyle \frac{1}{x^2} +4x +13 $?
Originally Posted by mindblowing ∫ (1/X2 +4X+13) dX Are you familiar with the standard form $\displaystyle \int \frac{a}{x^2 + a^2} \, dx = \tan^{-1} \left( \frac{x}{a}\right) + C$ ? Note that $\displaystyle \frac{1}{x^2 + 4x + 13} = \frac{1}{(x + 2)^2 + 9} = \frac{1}{3} \, \left( \frac{3}{(x + 2)^2 + 3^2}\right)$.
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