What is the integral of (x+1)/(x^4+3x^2) ?

The integral of (x+1)/(x^4+3x^2) is 1/3 ln(x)-1/6 ln|x^2+3|-1/(3x)-1/(3sqrt(3))arctan(x/(sqrt(3)))+C

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integral of (x+1)/(x^4+3x^2)

\int \frac{x + 1}{x ^{4} + 3 x ^{2}} d x

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

Solution steps

\int \frac{x + 1}{x ^{4} + 3 x ^{2}} d x

Expand

\frac{x + 1}{x ^{4} + 3 x ^{2}} : \frac{x}{x ^{4} + 3 x ^{2}} + \frac{1}{x ^{4} + 3 x ^{2}}

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}{x ^{4} + 3 x ^{2}} d x + \int \frac{1}{x ^{4} + 3 x ^{2}} d x

\int \frac{x}{x ^{4} + 3 x ^{2}} d x = \frac{1}{3} ln (x) - \frac{1}{6} ln x^{2} + 3

\int \frac{1}{x ^{4} + 3 x ^{2}} d x = - \frac{1}{3 x} - \frac{1}{3 3} arctan (\frac{x}{3})

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

Add a constant to the solution

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

What is the integral of (x+1)/(x^4+3x^2) ?
The integral of (x+1)/(x^4+3x^2) is 1/3 ln(x)-1/6 ln|x^2+3|-1/(3x)-1/(3sqrt(3))arctan(x/(sqrt(3)))+C