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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

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

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

Simplify

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

Add a constant to the solution

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

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