What is the integral of (x^4+3x^3+x+1)/(x^2+1) ?

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

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integral of (x^4+3x^3+x+1)/(x^2+1)

\int \frac{x ^{4} + 3 x ^{3} + x + 1}{x ^{2} + 1} d x

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

Solution steps

\int \frac{x ^{4} + 3 x ^{3} + x + 1}{x ^{2} + 1} d x

Expand

\frac{x ^{4} + 3 x ^{3} + x + 1}{x ^{2} + 1} : \frac{x ^{4}}{x ^{2} + 1} + \frac{3 x ^{3}}{x ^{2} + 1} + \frac{x}{x ^{2} + 1} + \frac{1}{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 ^{4}}{x ^{2} + 1} d x + \int \frac{3 x ^{3}}{x ^{2} + 1} d x + \int \frac{x}{x ^{2} + 1} d x + \int \frac{1}{x ^{2} + 1} d x

\int \frac{x ^{4}}{x ^{2} + 1} d x = \frac{x ^{3}}{3} - x + arctan (x)

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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