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

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

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

integral

\frac{4 x ^{4}}{( 1 + x ^{2} )}

4 (\frac{x ^{3}}{3} - x + arctan (x)) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Long division

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

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

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

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

\int 1 d x = x

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

= 4 (\frac{x ^{3}}{3} - x + arctan (x))

Add a constant to the solution

= 4 (\frac{x ^{3}}{3} - x + arctan (x)) + C

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