What is the integral of (x^2+1)/(2x^3+6x) ?

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

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integral of (x^2+1)/(2x^3+6x)

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

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

Solution steps

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

Factor

\frac{x ^{2} + 1}{2 x ^{3} + 6 x}

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{x ^{2} + 1}{x ( x ^{2} + 3 )} d x

Expand

\frac{x ^{2} + 1}{x ( x ^{2} + 3 )} : \frac{x ^{2}}{x ( x ^{2} + 3 )} + \frac{1}{x ( x ^{2} + 3 )}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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