What is the integral of (7-x^2)/(5x^3+x) ?

The integral of (7-x^2)/(5x^3+x) is 7ln|x|-18/5 ln|5x^2+1|+C

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integral of (7-x^2)/(5x^3+x)

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

7 ln ∣ x ∣ - \frac{18}{5} ln 5 x^{2} + 1 + C

Solution steps

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

Expand

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

\int \frac{7}{5 x ^{3} + x} d x = 7 (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

= 7 (ln ∣ x ∣ - \frac{1}{2} ln 5 x^{2} + 1) - \frac{1}{10} ln 5 x^{2} + 1

Simplify

7 (ln ∣ x ∣ - \frac{1}{2} ln 5 x^{2} + 1) - \frac{1}{10} ln 5 x^{2} + 1 : 7 ln ∣ x ∣ - \frac{18}{5} ln 5 x^{2} + 1

= 7 ln ∣ x ∣ - \frac{18}{5} ln 5 x^{2} + 1

Add a constant to the solution

= 7 ln ∣ x ∣ - \frac{18}{5} ln 5 x^{2} + 1 + C

x \to 0 lim (x e^{- \frac{1}{x}})

f (x) = 8 x, at x = 16

lo g_{3} (2 x^{2} + 4 x + 3)

n = 1 \sum \infty \frac{9 9 ^{n} ( n ^{5} + 2 )}{n ^{2} 1 0 ^{2 n}}

f (x) = \frac{11}{ln ( x )}

What is the integral of (7-x^2)/(5x^3+x) ?
The integral of (7-x^2)/(5x^3+x) is 7ln|x|-18/5 ln|5x^2+1|+C