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

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

n = 1 \sum \infty \frac{1}{n ^{5}}

f (x) = (12 - x) 24 x - 144

y^{^{'''}} - y^{^{''}} + 6 y^{^{'}} - 6 y = 0

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

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