What is the integral of (x^3)/(x^4-6x^2+1) ?

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

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integral of (x^3)/(x^4-6x^2+1)

\int \frac{x ^{3}}{x ^{4} - 6 x ^{2} + 1} d x

\frac{1}{2} (\frac{1}{2} ln x^{4} - 6 x^{2} + 1 - \frac{3}{4 2} (ln \frac{1}{2 2} (x^{2} - 3) + 1 - ln \frac{1}{2 2} (x^{2} - 3) - 1)) + C

Solution steps

\int \frac{x ^{3}}{x ^{4} - 6 x ^{2} + 1} d x

Apply u-substitution

: \int \frac{u}{2 ( u ^{2} + 1 - 6 u )} d u

= \int \frac{u}{2 ( u ^{2} + 1 - 6 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}{u ^{2} + 1 - 6 u} d u

Complete the square

u^{2} + 1 - 6 u : (u - 3)^{2} - 8

= \frac{1}{2} \cdot \int \frac{u}{( u - 3 ) ^{2} - 8} d u

Apply u-substitution

: \int \frac{v + 3}{v ^{2} - 8} d v

= \frac{1}{2} \cdot \int \frac{v + 3}{v ^{2} - 8} d v

Expand

\frac{v + 3}{v ^{2} - 8} : \frac{v}{v ^{2} - 8} + \frac{3}{v ^{2} - 8}

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 \frac{v}{v ^{2} - 8} d v + \int \frac{3}{v ^{2} - 8} d v)

\int \frac{v}{v ^{2} - 8} d v = \frac{1}{2} ln v^{2} - 8

\int \frac{3}{v ^{2} - 8} d v = - \frac{3}{4 2} (ln \frac{v}{2 2} + 1 - ln \frac{v}{2 2} - 1)

= \frac{1}{2} (\frac{1}{2} ln v^{2} - 8 - \frac{3}{4 2} (ln \frac{v}{2 2} + 1 - ln \frac{v}{2 2} - 1))

Substitute back

= \frac{1}{2} (\frac{1}{2} ln (x^{2} - 3)^{2} - 8 - \frac{3}{4 2} (ln \frac{x ^{2} - 3}{2 2} + 1 - ln \frac{x ^{2} - 3}{2 2} - 1))

Simplify

\frac{1}{2} (\frac{1}{2} ln (x^{2} - 3)^{2} - 8 - \frac{3}{4 2} (ln \frac{x ^{2} - 3}{2 2} + 1 - ln \frac{x ^{2} - 3}{2 2} - 1)) : \frac{1}{2} (\frac{1}{2} ln x^{4} - 6 x^{2} + 1 - \frac{3}{4 2} (ln \frac{1}{2 2} (x^{2} - 3) + 1 - ln \frac{1}{2 2} (x^{2} - 3) - 1))

= \frac{1}{2} (\frac{1}{2} ln x^{4} - 6 x^{2} + 1 - \frac{3}{4 2} (ln \frac{1}{2 2} (x^{2} - 3) + 1 - ln \frac{1}{2 2} (x^{2} - 3) - 1))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{2} ln x^{4} - 6 x^{2} + 1 - \frac{3}{4 2} (ln \frac{1}{2 2} (x^{2} - 3) + 1 - ln \frac{1}{2 2} (x^{2} - 3) - 1)) + C

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