What is the integral of (9x^2+x+48)/((x+3)(x^2+9)) ?

The integral of (9x^2+x+48)/((x+3)(x^2+9)) is 7ln|x+3|+ln|(x^2)/9+1|-5/3 arctan(x/3)+C

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integral of (9x^2+x+48)/((x+3)(x^2+9))

\int \frac{9 x ^{2} + x + 48}{( x + 3 ) ( x ^{2} + 9 )} d x

7 ln ∣ x + 3 ∣ + ln \frac{x ^{2}}{9} + 1 - \frac{5}{3} arctan (\frac{x}{3}) + C

Solution steps

\int \frac{9 x ^{2} + x + 48}{( x + 3 ) ( x ^{2} + 9 )} d x

Take the partial fraction of

\frac{9 x ^{2} + x + 48}{( x + 3 ) ( x ^{2} + 9 )} : \frac{7}{x + 3} + \frac{2 x - 5}{x ^{2} + 9}

= \int \frac{7}{x + 3} + \frac{2 x - 5}{x ^{2} + 9} 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{7}{x + 3} d x + \int \frac{2 x - 5}{x ^{2} + 9} d x

\int \frac{7}{x + 3} d x = 7 ln ∣ x + 3 ∣

\int \frac{2 x - 5}{x ^{2} + 9} d x = ln \frac{x ^{2}}{9} + 1 - \frac{5}{3} arctan (\frac{x}{3})

= 7 ln ∣ x + 3 ∣ + ln \frac{x ^{2}}{9} + 1 - \frac{5}{3} arctan (\frac{x}{3})

Add a constant to the solution

= 7 ln ∣ x + 3 ∣ + ln \frac{x ^{2}}{9} + 1 - \frac{5}{3} arctan (\frac{x}{3}) + C

\frac{\partial}{\partial y} (x y e^{s i n (π x y)})

y = 2, y = sec (x)

\frac{d}{d t} (sin^{2} (π t))

\frac{d}{d x} (\frac{1}{x ^{2} + y ^{2} + z ^{2}})

\frac{\partial}{\partial x} (x e^{x^{2} y})

What is the integral of (9x^2+x+48)/((x+3)(x^2+9)) ?
The integral of (9x^2+x+48)/((x+3)(x^2+9)) is 7ln|x+3|+ln|(x^2)/9+1|-5/3 arctan(x/3)+C