What is the integral of (x^4)/(x^2+9) ?

The integral of (x^4)/(x^2+9) is 27((x^3)/(81)-x/3+arctan(x/3))+C

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integral of (x^4)/(x^2+9)

\int \frac{x ^{4}}{x ^{2} + 9} d x

27 (\frac{x ^{3}}{81} - \frac{x}{3} + arctan (\frac{x}{3})) + C

Solution steps

\int \frac{x ^{4}}{x ^{2} + 9} d x

Apply Integral Substitution

= \int \frac{27 u ^{4}}{u ^{2} + 1} d u

Take the constant out:

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

= 27 \cdot \int \frac{u ^{4}}{u ^{2} + 1} d u

Long division

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

= 27 \cdot \int u^{2} - 1 + \frac{1}{u ^{2} + 1} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 27 (\int u^{2} d u - \int 1 d u + \int \frac{1}{u ^{2} + 1} d u)

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

\int 1 d u = u

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

= 27 (\frac{u ^{3}}{3} - u + arctan (u))

Substitute back

u = \frac{x}{3}

= 27 (\frac{( \frac{x}{3} ) ^{3}}{3} - \frac{x}{3} + arctan (\frac{x}{3}))

Simplify

27 (\frac{( \frac{x}{3} ) ^{3}}{3} - \frac{x}{3} + arctan (\frac{x}{3})) : 27 (\frac{x ^{3}}{81} - \frac{x}{3} + arctan (\frac{x}{3}))

= 27 (\frac{x ^{3}}{81} - \frac{x}{3} + arctan (\frac{x}{3}))

Add a constant to the solution

= 27 (\frac{x ^{3}}{81} - \frac{x}{3} + arctan (\frac{x}{3})) + C

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What is the integral of (x^4)/(x^2+9) ?
The integral of (x^4)/(x^2+9) is 27((x^3)/(81)-x/3+arctan(x/3))+C