What is the integral of (9-x^2)/(5x^3+x) ?

The integral of (9-x^2)/(5x^3+x) is 9ln|x|-23/5 ln|5x^2+1|+C

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integral of (9-x^2)/(5x^3+x)

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

9 ln ∣ x ∣ - \frac{23}{5} ln 5 x^{2} + 1 + C

Solution steps

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

Expand

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

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

\int \frac{x ^{2}}{5 x ^{3} + x} d x = \frac{1}{10} ln 5 x^{2} + 1

= 9 (ln ∣ x ∣ - \frac{1}{2} ln 5 x^{2} + 1) - \frac{1}{10} ln 5 x^{2} + 1

Simplify

9 (ln ∣ x ∣ - \frac{1}{2} ln 5 x^{2} + 1) - \frac{1}{10} ln 5 x^{2} + 1 : 9 ln ∣ x ∣ - \frac{23}{5} ln 5 x^{2} + 1

= 9 ln ∣ x ∣ - \frac{23}{5} ln 5 x^{2} + 1

Add a constant to the solution

= 9 ln ∣ x ∣ - \frac{23}{5} ln 5 x^{2} + 1 + C

f (x) = (\frac{2 x + 1}{x + 5}) (3 x - 1)

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What is the integral of (9-x^2)/(5x^3+x) ?
The integral of (9-x^2)/(5x^3+x) is 9ln|x|-23/5 ln|5x^2+1|+C