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

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

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

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

- \frac{1}{4} ln - 4 x^{4} - 4 x^{2} + 4 + \frac{1}{4 5} (ln \frac{2 x ^{2} + 1}{5} + 1 - ln \frac{2 x ^{2} + 1}{5} - 1) + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{v}{- v ^{2} + v + 1} d v

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

Complete the square

- v^{2} + v + 1 : - (v - \frac{1}{2})^{2} + \frac{5}{4}

= \frac{1}{2} \cdot \int \frac{v}{- ( v - \frac{1}{2} ) ^{2} + \frac{5}{4}} d v

Apply u-substitution

: \int \frac{2 ( 2 w + 1 )}{- 4 w ^{2} + 5} d w

= \frac{1}{2} \cdot \int \frac{2 ( 2 w + 1 )}{- 4 w ^{2} + 5} d w

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int \frac{2 w + 1}{- 4 w ^{2} + 5} d w

Expand

\frac{2 w + 1}{- 4 w ^{2} + 5} : \frac{2 w}{- 4 w ^{2} + 5} + \frac{1}{- 4 w ^{2} + 5}

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} \cdot 2 (\int \frac{2 w}{- 4 w ^{2} + 5} d w + \int \frac{1}{- 4 w ^{2} + 5} d w)

\int \frac{2 w}{- 4 w ^{2} + 5} d w = - \frac{1}{4} ln - 4 w^{2} + 5

\int \frac{1}{- 4 w ^{2} + 5} d w = \frac{1}{4 5} (ln \frac{2}{5} w + 1 - ln \frac{2}{5} w - 1)

= \frac{1}{2} \cdot 2 (- \frac{1}{4} ln - 4 w^{2} + 5 + \frac{1}{4 5} (ln \frac{2}{5} w + 1 - ln \frac{2}{5} w - 1))

Substitute back

= \frac{1}{2} \cdot 2 (- \frac{1}{4} ln - 4 (x^{2} + 1 - \frac{1}{2})^{2} + 5 + \frac{1}{4 5} (ln \frac{2}{5} (x^{2} + 1 - \frac{1}{2}) + 1 - ln \frac{2}{5} (x^{2} + 1 - \frac{1}{2}) - 1))

Simplify

\frac{1}{2} \cdot 2 (- \frac{1}{4} ln - 4 (x^{2} + 1 - \frac{1}{2})^{2} + 5 + \frac{1}{4 5} (ln \frac{2}{5} (x^{2} + 1 - \frac{1}{2}) + 1 - ln \frac{2}{5} (x^{2} + 1 - \frac{1}{2}) - 1)) : - \frac{1}{4} ln - 4 x^{4} - 4 x^{2} + 4 + \frac{1}{4 5} (ln \frac{2 x ^{2} + 1}{5} + 1 - ln \frac{2 x ^{2} + 1}{5} - 1)

= - \frac{1}{4} ln - 4 x^{4} - 4 x^{2} + 4 + \frac{1}{4 5} (ln \frac{2 x ^{2} + 1}{5} + 1 - ln \frac{2 x ^{2} + 1}{5} - 1)

Add a constant to the solution

= - \frac{1}{4} ln - 4 x^{4} - 4 x^{2} + 4 + \frac{1}{4 5} (ln \frac{2 x ^{2} + 1}{5} + 1 - ln \frac{2 x ^{2} + 1}{5} - 1) + C

x \to 2 lim (2 x^{2} - 4 x + 3)

\int \frac{e ^{y} sin ( 2 x )}{cos ( x )} d x

3 x sin (x)

ln (x^{8})

\frac{d}{d x} ((\frac{x ^{2} + 4}{x ^{2} - 4})^{5})

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