What is the integral of (x+9)/(x^2-6x+9) ?

The integral of (x+9)/(x^2-6x+9) is ln|x-3|-(12)/(x-3)+C

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integral of (x+9)/(x^2-6x+9)

\int \frac{x + 9}{x ^{2} - 6 x + 9} d x

ln ∣ x - 3 ∣ - \frac{12}{x - 3} + C

Solution steps

\int \frac{x + 9}{x ^{2} - 6 x + 9} d x

Take the partial fraction of

\frac{x + 9}{x ^{2} - 6 x + 9} : \frac{1}{x - 3} + \frac{12}{( x - 3 ) ^{2}}

= \int \frac{1}{x - 3} + \frac{12}{( x - 3 ) ^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{x - 3} d x + \int \frac{12}{( x - 3 ) ^{2}} d x

\int \frac{1}{x - 3} d x = ln ∣ x - 3 ∣

\int \frac{12}{( x - 3 ) ^{2}} d x = - \frac{12}{x - 3}

= ln ∣ x - 3 ∣ - \frac{12}{x - 3}

Add a constant to the solution

= ln ∣ x - 3 ∣ - \frac{12}{x - 3} + C

y (\frac{π}{21}) = 2 tan^{2} (7 x)

\frac{d}{d x} (2^{\frac{x}{2}})

\frac{d}{d x} (e^{\frac{- x}{2}})

\int 15 e^{x} d x

\frac{\partial}{\partial u ( v )} (e^{2 u (v) v})

What is the integral of (x+9)/(x^2-6x+9) ?
The integral of (x+9)/(x^2-6x+9) is ln|x-3|-(12)/(x-3)+C