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

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

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

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

x^{3} + arctan (x) + C

Solution steps

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

Expand

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

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

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

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

= x^{3} - 3 x + 3 arctan (x) + 3 (- arctan (x) + x) + arctan (x)

Simplify

x^{3} - 3 x + 3 arctan (x) + 3 (- arctan (x) + x) + arctan (x) : x^{3} + arctan (x)

= x^{3} + arctan (x)

Add a constant to the solution

= x^{3} + arctan (x) + C

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sin^{3} (x)

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