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

The integral of (7-x^2)/(3x^3+x) is 7ln|x|-11/3 ln|3x^2+1|+C

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

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

7 ln ∣ x ∣ - \frac{11}{3} ln 3 x^{2} + 1 + C

Solution steps

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

Expand

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

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

\int \frac{x ^{2}}{3 x ^{3} + x} d x = \frac{1}{6} ln 3 x^{2} + 1

= 7 (ln ∣ x ∣ - \frac{1}{2} ln 3 x^{2} + 1) - \frac{1}{6} ln 3 x^{2} + 1

Simplify

7 (ln ∣ x ∣ - \frac{1}{2} ln 3 x^{2} + 1) - \frac{1}{6} ln 3 x^{2} + 1 : 7 ln ∣ x ∣ - \frac{11}{3} ln 3 x^{2} + 1

= 7 ln ∣ x ∣ - \frac{11}{3} ln 3 x^{2} + 1

Add a constant to the solution

= 7 ln ∣ x ∣ - \frac{11}{3} ln 3 x^{2} + 1 + C

\frac{\partial}{\partial y} (x^{5} y z^{2} + 4 yz)

f (x) = x^{2} + x, at a = 5

\frac{d}{d x} (I n x y (I, x)^{2})

\int \frac{13}{x ( x ^{2} + 4 ) ^{2}} d x

y = 38 x - 16 x^{2}, at x = 1

What is the integral of (7-x^2)/(3x^3+x) ?
The integral of (7-x^2)/(3x^3+x) is 7ln|x|-11/3 ln|3x^2+1|+C