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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

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

\int_{1}^{2} \frac{x ^{2} + 1}{x ^{2}} d x

sin (t + 1)

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