What is the integral of (x^2)/(1-x^6) ?

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

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integral of (x^2)/(1-x^6)

\int \frac{x ^{2}}{1 - x ^{6}} d x

\frac{1}{6} (ln x^{3} + 1 - ln x^{3} - 1) + C

Solution steps

\int \frac{x ^{2}}{1 - x ^{6}} d x

Apply u-substitution

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

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

Use the common integral:

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

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

Substitute back

u = x^{3}

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

Simplify

\frac{1}{3} (\frac{ln x ^{3} + 1}{2} - \frac{ln x ^{3} - 1}{2}) : \frac{1}{6} (ln x^{3} + 1 - ln x^{3} - 1)

= \frac{1}{6} (ln x^{3} + 1 - ln x^{3} - 1)

Add a constant to the solution

= \frac{1}{6} (ln x^{3} + 1 - ln x^{3} - 1) + C

\int \frac{2 x - 5}{( x + 2 ) ( x + 3 )} d x

\int x cos (x) d x

\frac{3 x ^{\frac{1}{2}}}{2}

x \to - 7 lim (\frac{6 x + 42}{x + 7})

\frac{d}{d x} (u (x) (x)^{n})

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