What is the integral of (x^3)/(x^2-36) ?

The integral of (x^3)/(x^2-36) is 1/2 (x^2-36+36ln|x^2-36|)+C

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integral of (x^3)/(x^2-36)

\int \frac{x ^{3}}{x ^{2} - 36} d x

\frac{1}{2} (x^{2} - 36 + 36 ln x^{2} - 36) + C

Solution steps

\int \frac{x ^{3}}{x ^{2} - 36} d x

Apply u-substitution

: \int \frac{u + 36}{2 u} d u

= \int \frac{u + 36}{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 + 36}{u} d u

Expand

\frac{u + 36}{u} : 1 + \frac{36}{u}

= \frac{1}{2} \cdot \int 1 + \frac{36}{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{36}{u} d u)

\int 1 d u = u

\int \frac{36}{u} d u = 36 ln ∣ u ∣

= \frac{1}{2} (u + 36 ln ∣ u ∣)

Substitute back

u = x^{2} - 36

= \frac{1}{2} (x^{2} - 36 + 36 ln x^{2} - 36)

Add a constant to the solution

= \frac{1}{2} (x^{2} - 36 + 36 ln x^{2} - 36) + C

\frac{\partial}{\partial x} (e^{2 x} \cdot cos (x + y))

y = ∣ 2 x ∣, y = x^{2} - 3

\frac{d}{d x} (e^{- 2 x} (sin (3 x) + cos (3 x)))

\frac{d}{d x} (- 8 sin (4 x))

x \to 3 + lim (x)

What is the integral of (x^3)/(x^2-36) ?
The integral of (x^3)/(x^2-36) is 1/2 (x^2-36+36ln|x^2-36|)+C