What is the integral of (-x^3+7x^2-x)/(x^2+1) ?

The integral of (-x^3+7x^2-x)/(x^2+1) is -1/2 x^2+7x-7arctan(x)-1/2+C

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integral of (-x^3+7x^2-x)/(x^2+1)

\int \frac{- x ^{3} + 7 x ^{2} - x}{x ^{2} + 1} d x

- \frac{1}{2} x^{2} + 7 x - 7 arctan (x) - \frac{1}{2} + C

Solution steps

\int \frac{- x ^{3} + 7 x ^{2} - x}{x ^{2} + 1} d x

Expand

\frac{- x ^{3} + 7 x ^{2} - x}{x ^{2} + 1} : - \frac{x ^{3}}{x ^{2} + 1} + \frac{7 x ^{2}}{x ^{2} + 1} - \frac{x}{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 ^{3}}{x ^{2} + 1} d x + \int \frac{7 x ^{2}}{x ^{2} + 1} d x - \int \frac{x}{x ^{2} + 1} d x

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

\int \frac{7 x ^{2}}{x ^{2} + 1} d x = 7 (- arctan (x) + x)

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

= - \frac{1}{2} (x^{2} + 1 - ln x^{2} + 1) + 7 (- arctan (x) + x) - \frac{1}{2} ln x^{2} + 1

Simplify

- \frac{1}{2} (x^{2} + 1 - ln x^{2} + 1) + 7 (- arctan (x) + x) - \frac{1}{2} ln x^{2} + 1 : - \frac{1}{2} x^{2} + 7 x - 7 arctan (x) - \frac{1}{2}

= - \frac{1}{2} x^{2} + 7 x - 7 arctan (x) - \frac{1}{2}

Add a constant to the solution

= - \frac{1}{2} x^{2} + 7 x - 7 arctan (x) - \frac{1}{2} + C

What is the integral of (-x^3+7x^2-x)/(x^2+1) ?
The integral of (-x^3+7x^2-x)/(x^2+1) is -1/2 x^2+7x-7arctan(x)-1/2+C