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

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

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

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

\frac{x ^{3}}{3} - x + \frac{1}{2} ln x^{2} + 1 + C

Solution steps

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

Long division

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

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

\int x^{2} d x = \frac{x ^{3}}{3}

\int 1 d x = x

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

= \frac{x ^{3}}{3} - x + \frac{1}{2} ln x^{2} + 1

Add a constant to the solution

= \frac{x ^{3}}{3} - x + \frac{1}{2} ln x^{2} + 1 + C

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

\frac{\partial}{\partial x} ((1 - x) + 10 (y - x^{2})^{2})

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

\frac{d}{d x} (x - 2 - 0.5 sin (x))

\frac{d}{d s} (s t)

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