What is the integral of (7x^2+x+54)/(x^3+9x) ?

The integral of (7x^2+x+54)/(x^3+9x) is 6ln|x|+1/2 ln|x^2+9|+1/3 arctan(x/3)+C

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integral of (7x^2+x+54)/(x^3+9x)

\int \frac{7 x ^{2} + x + 54}{x ^{3} + 9 x} d x

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

Solution steps

\int \frac{7 x ^{2} + x + 54}{x ^{3} + 9 x} d x

Take the partial fraction of

\frac{7 x ^{2} + x + 54}{x ^{3} + 9 x} : \frac{6}{x} + \frac{x + 1}{x ^{2} + 9}

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

\int \frac{6}{x} d x = 6 ln ∣ x ∣

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

= 6 ln ∣ x ∣ + \frac{1}{2} ln x^{2} + 9 + \frac{1}{3} arctan (\frac{x}{3})

Add a constant to the solution

= 6 ln ∣ x ∣ + \frac{1}{2} ln x^{2} + 9 + \frac{1}{3} arctan (\frac{x}{3}) + C

y = x + 2

\frac{d}{d x} (\frac{5}{25 x ^{2} + 1})

\int \frac{2 t ^{5}}{5} - \frac{2 t ^{2}}{5} - \frac{2 t}{5} d t

\frac{d}{d x} (ln (3^{3 x}))

\int_{0}^{\frac{1}{2}} 5 cos (- 1 (x)) d x

What is the integral of (7x^2+x+54)/(x^3+9x) ?
The integral of (7x^2+x+54)/(x^3+9x) is 6ln|x|+1/2 ln|x^2+9|+1/3 arctan(x/3)+C