What is the integral of (x^3)/(x^2-81) ?

The integral of (x^3)/(x^2-81) is 1/2 (x^2-81+81ln|x^2-81|)+C

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integral of (x^3)/(x^2-81)

\int \frac{x ^{3}}{x ^{2} - 81} d x

\frac{1}{2} (x^{2} - 81 + 81 ln x^{2} - 81) + C

Solution steps

\int \frac{x ^{3}}{x ^{2} - 81} d x

Apply u-substitution

: \int \frac{u + 81}{2 u} d u

= \int \frac{u + 81}{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 + 81}{u} d u

Expand

\frac{u + 81}{u} : 1 + \frac{81}{u}

= \frac{1}{2} \cdot \int 1 + \frac{81}{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{81}{u} d u)

\int 1 d u = u

\int \frac{81}{u} d u = 81 ln ∣ u ∣

= \frac{1}{2} (u + 81 ln ∣ u ∣)

Substitute back

u = x^{2} - 81

= \frac{1}{2} (x^{2} - 81 + 81 ln x^{2} - 81)

Add a constant to the solution

= \frac{1}{2} (x^{2} - 81 + 81 ln x^{2} - 81) + C

y (θ) = \frac{sin ( θ )}{5} + \frac{c}{θ}

\int (\frac{x - 1}{x})^{2} d x

\frac{d}{d x} (3 (cos (x))^{2})

n = 1 \sum \infty \frac{1}{2 ^{n} + 1}

\frac{d}{d x} (\frac{1}{e ^{x^{2}}})

What is the integral of (x^3)/(x^2-81) ?
The integral of (x^3)/(x^2-81) is 1/2 (x^2-81+81ln|x^2-81|)+C