What is the integral of (x^5)/(1-x^4) ?

The integral of (x^5)/(1-x^4) is -(x^2)/2+1/4 (ln|x^2+1|-ln|x^2-1|)+C

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integral of (x^5)/(1-x^4)

\int \frac{x ^{5}}{1 - x ^{4}} d x

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

Solution steps

\int \frac{x ^{5}}{1 - x ^{4}} d x

Long division

\frac{x ^{5}}{1 - x ^{4}} : - x + \frac{x}{- x ^{4} + 1}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int x d x = \frac{x ^{2}}{2}

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

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

Add a constant to the solution

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

\int (5 x + 5) (x - 4) (x - 4) d x

\int 9 t^{2} + \frac{1}{t ^{3}} d t

y^{^{''}} + y^{^{'}} - 2 y = 6 e^{x}

4 y^{^{''}} + 8 y^{^{'}} + 3 y = 0

\frac{5 s}{( s ^{2} + 4 ) \cdot ( s + 7 )}

What is the integral of (x^5)/(1-x^4) ?
The integral of (x^5)/(1-x^4) is -(x^2)/2+1/4 (ln|x^2+1|-ln|x^2-1|)+C