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

The integral of (6-2x)/(9-x^2) is ln| x/3+1|-ln| x/3-1|+ln|9-x^2|+C

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

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

ln \frac{x}{3} + 1 - ln \frac{x}{3} - 1 + ln 9 - x^{2} + C

Solution steps

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

Expand

\frac{6 - 2 x}{9 - x ^{2}} : \frac{6}{9 - x ^{2}} - \frac{2 x}{9 - x ^{2}}

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

\int \frac{6}{9 - x ^{2}} d x = ln \frac{x}{3} + 1 - ln \frac{x}{3} - 1

\int \frac{2 x}{9 - x ^{2}} d x = - ln 9 - x^{2}

= ln \frac{x}{3} + 1 - ln \frac{x}{3} - 1 - (- ln 9 - x^{2})

Simplify

= ln \frac{x}{3} + 1 - ln \frac{x}{3} - 1 + ln 9 - x^{2}

Add a constant to the solution

= ln \frac{x}{3} + 1 - ln \frac{x}{3} - 1 + ln 9 - x^{2} + C

\int \frac{5 x - 6}{x ^{2} + 4} d x

\frac{\partial}{\partial x} (sin (7 x^{4} y - 7 x y^{4}))

y^{^{''}} - 10 y^{^{'}} + 34 y = 0

\frac{\partial}{\partial x} (x^{- \frac{1}{2}} cos (x))

\frac{s - 1}{s ( s ^{2} + s + 1 )}

What is the integral of (6-2x)/(9-x^2) ?
The integral of (6-2x)/(9-x^2) is ln| x/3+1|-ln| x/3-1|+ln|9-x^2|+C