What is the integral of (x^2)/(x^6-4) ?

The integral of (x^2)/(x^6-4) is -1/12 (ln|(x^3)/2+1|-ln|(x^3)/2-1|)+C

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integral of (x^2)/(x^6-4)

\int \frac{x ^{2}}{x ^{6} - 4} d x

- \frac{1}{12} (ln \frac{x ^{3}}{2} + 1 - ln \frac{x ^{3}}{2} - 1) + C

Solution steps

\int \frac{x ^{2}}{x ^{6} - 4} d x

Apply u-substitution

: \int \frac{1}{3 ( u ^{2} - 4 )} d u

= \int \frac{1}{3 ( u ^{2} - 4 )} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{u ^{2} - 4} d u

Apply Integral Substitution

= \frac{1}{3} \cdot \int \frac{1}{2 ( v ^{2} - 1 )} d v

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

v^{2} - 1 : - (- v^{2} + 1)

= \frac{1}{3} \cdot \frac{1}{2} \cdot \int \frac{1}{- ( - v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{2} (- \int \frac{1}{- v ^{2} + 1} d v)

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= \frac{1}{3} \cdot \frac{1}{2} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= \frac{1}{3} \cdot \frac{1}{2} - \frac{ln \frac{x ^{3}}{2} + 1}{2} - \frac{ln \frac{x ^{3}}{2} - 1}{2}

Simplify

\frac{1}{3} \cdot \frac{1}{2} - \frac{ln \frac{x ^{3}}{2} + 1}{2} - \frac{ln \frac{x ^{3}}{2} - 1}{2} : - \frac{1}{12} (ln \frac{x ^{3}}{2} + 1 - ln \frac{x ^{3}}{2} - 1)

= - \frac{1}{12} (ln \frac{x ^{3}}{2} + 1 - ln \frac{x ^{3}}{2} - 1)

Add a constant to the solution

= - \frac{1}{12} (ln \frac{x ^{3}}{2} + 1 - ln \frac{x ^{3}}{2} - 1) + C

\frac{\partial}{\partial x} (x e^{6 x y})

\int_{0}^{\infty} \frac{2}{x ^{2} + 4} d x

(1 + x^{2}) \frac{d y}{d x} + 3 x y = 5 x

y^{2} \frac{d y}{d x} = 1

\int x arctan (20 x) d x

What is the integral of (x^2)/(x^6-4) ?
The integral of (x^2)/(x^6-4) is -1/12 (ln|(x^3)/2+1|-ln|(x^3)/2-1|)+C