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

The integral of (9-x^2)/(7x^3+x) is 9ln|x|-32/7 ln|7x^2+1|+C

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

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

9 ln ∣ x ∣ - \frac{32}{7} ln 7 x^{2} + 1 + C

Solution steps

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

Expand

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

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

\int \frac{x ^{2}}{7 x ^{3} + x} d x = \frac{1}{14} ln 7 x^{2} + 1

= 9 (ln ∣ x ∣ - \frac{1}{2} ln 7 x^{2} + 1) - \frac{1}{14} ln 7 x^{2} + 1

Simplify

9 (ln ∣ x ∣ - \frac{1}{2} ln 7 x^{2} + 1) - \frac{1}{14} ln 7 x^{2} + 1 : 9 ln ∣ x ∣ - \frac{32}{7} ln 7 x^{2} + 1

= 9 ln ∣ x ∣ - \frac{32}{7} ln 7 x^{2} + 1

Add a constant to the solution

= 9 ln ∣ x ∣ - \frac{32}{7} ln 7 x^{2} + 1 + C

x \to 2 lim (\frac{x ^{7} - 128}{x - 2})

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

\int - 2 x y d x

\int y - 2 d y

n = 1 \sum \infty \frac{1}{n ^{0.9}}

What is the integral of (9-x^2)/(7x^3+x) ?
The integral of (9-x^2)/(7x^3+x) is 9ln|x|-32/7 ln|7x^2+1|+C