What is the integral of (x^3)/(x^2-16) ?

The integral of (x^3)/(x^2-16) is 1/2 (x^2-16+16ln|x^2-16|)+C

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integral of (x^3)/(x^2-16)

\int \frac{x ^{3}}{x ^{2} - 16} d x

\frac{1}{2} (x^{2} - 16 + 16 ln x^{2} - 16) + C

Solution steps

\int \frac{x ^{3}}{x ^{2} - 16} d x

Apply u-substitution

: \int \frac{u + 16}{2 u} d u

= \int \frac{u + 16}{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 + 16}{u} d u

Expand

\frac{u + 16}{u} : 1 + \frac{16}{u}

= \frac{1}{2} \cdot \int 1 + \frac{16}{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{16}{u} d u)

\int 1 d u = u

\int \frac{16}{u} d u = 16 ln ∣ u ∣

= \frac{1}{2} (u + 16 ln ∣ u ∣)

Substitute back

u = x^{2} - 16

= \frac{1}{2} (x^{2} - 16 + 16 ln x^{2} - 16)

Add a constant to the solution

= \frac{1}{2} (x^{2} - 16 + 16 ln x^{2} - 16) + C

x \to \infty lim (4 x^{4})

\frac{d y}{d x} = tan (x + y)

\frac{d}{d x} (3 (0)^{2} - 3)

x \to 1 - lim (- \frac{1}{x - 2})

\frac{1}{( s - 1 ) ( s )}

What is the integral of (x^3)/(x^2-16) ?
The integral of (x^3)/(x^2-16) is 1/2 (x^2-16+16ln|x^2-16|)+C