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

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

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

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

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

Solution steps

\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

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

Add similar elements:

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

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

Add a constant to the solution

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

\frac{1}{x} cos (x)

\frac{s + 5}{7 s ^{2} + 26 s + 25}

\frac{d y}{d t} + 4 y = 0

f (x) = x^{7}, at x = 2

\frac{d v}{d t} = - k t v^{2}, v (0) = 50

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