What is the integral of (2*x^2-x+2)/(x^5+2*x^3+x) ?

The integral of (2*x^2-x+2)/(x^5+2*x^3+x) is 2ln|x|-ln|x^2+1|-1/2 (arctan(x)+x/(1+x^2))+C

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integral of (2x^2-x+2)/(x^5+2x^3+x)

\int \frac{2 \cdot x ^{2} - x + 2}{x ^{5} + 2 \cdot x ^{3} + x} d x

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

Solution steps

\int \frac{2 x ^{2} - x + 2}{x ^{5} + 2 x ^{3} + x} d x

Take the partial fraction of

\frac{2 x ^{2} - x + 2}{x ^{5} + 2 x ^{3} + x} : \frac{2}{x} - \frac{2 x}{x ^{2} + 1} - \frac{1}{( x ^{2} + 1 ) ^{2}}

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

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

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

Add a constant to the solution

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

\int - x^{2} - 4 x d x

\int (tan (x))^{3} sec (x) d x

\int 84 d x

3^{s i n (3 x)}

\frac{\partial}{\partial x} (7 ln (\frac{2 x y}{5 z}))

What is the integral of (2*x^2-x+2)/(x^5+2*x^3+x) ?
The integral of (2*x^2-x+2)/(x^5+2*x^3+x) is 2ln|x|-ln|x^2+1|-1/2 (arctan(x)+x/(1+x^2))+C