What is the integral of (x^3)/(x^2+9) ?

The integral of (x^3)/(x^2+9) is 1/2 (x^2+9-9ln|x^2+9|)+C

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integral of (x^3)/(x^2+9)

\int \frac{x ^{3}}{x ^{2} + 9} d x

\frac{1}{2} (x^{2} + 9 - 9 ln x^{2} + 9) + C

Solution steps

\int \frac{x ^{3}}{x ^{2} + 9} d x

Apply u-substitution

: \int \frac{u - 9}{2 u} d u

= \int \frac{u - 9}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{u - 9}{u} d u

Expand

\frac{u - 9}{u} : 1 - \frac{9}{u}

= \frac{1}{2} \cdot \int 1 - \frac{9}{u} 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

= \frac{1}{2} (\int 1 d u - \int \frac{9}{u} d u)

\int 1 d u = u

\int \frac{9}{u} d u = 9 ln ∣ u ∣

= \frac{1}{2} (u - 9 ln ∣ u ∣)

Substitute back

u = x^{2} + 9

= \frac{1}{2} (x^{2} + 9 - 9 ln x^{2} + 9)

Add a constant to the solution

= \frac{1}{2} (x^{2} + 9 - 9 ln x^{2} + 9) + C

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\frac{d y}{d x}, 1 = (\frac{y}{x})

What is the integral of (x^3)/(x^2+9) ?
The integral of (x^3)/(x^2+9) is 1/2 (x^2+9-9ln|x^2+9|)+C