How do you integrate $\int \frac{1}{x^{2} + 6 x + 9} d x$ $int 1/(x^2+6x+9) dx$ using partial fractions?

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Answer 1 · 2016-05-31T14:47:33

It is

$\int \frac{1}{x^{2} + 6 x + 9} d x = \int \frac{1}{{(x + 3)}^{2}} d x = - \frac{1}{x + 3} + c$ $int 1/(x^2+6x+9) dx=int 1/(x+3)^2 dx=-1/(x+3)+c$

How do you integrate ∫1x2+6x+9dxint 1/(x^2+6x+9) dx using partial fractions?