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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

= x^{3} - 3 x + 3 arctan (x) + 2 (- arctan (x) + x)

Simplify

x^{3} - 3 x + 3 arctan (x) + 2 (- arctan (x) + x) : x^{3} - x + arctan (x)

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

Add a constant to the solution

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

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