What is the integral of 1/(x^8+x^6) ?

The integral of 1/(x^8+x^6) is -1/x+1/(3x^3)-1/(5x^5)-arctan(x)+C

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integral of 1/(x^8+x^6)

\int \frac{1}{x ^{8} + x ^{6}} d x

- \frac{1}{x} + \frac{1}{3 x ^{3}} - \frac{1}{5 x ^{5}} - arctan (x) + C

Solution steps

\int \frac{1}{x ^{8} + x ^{6}} d x

Take the partial fraction of

\frac{1}{x ^{8} + x ^{6}} : \frac{1}{x ^{2}} - \frac{1}{x ^{4}} + \frac{1}{x ^{6}} - \frac{1}{x ^{2} + 1}

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

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

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

\int \frac{1}{x ^{6}} d x = - \frac{1}{5 x ^{5}}

\int \frac{1}{x ^{2} + 1} d x = arctan (x)

= - \frac{1}{x} - (- \frac{1}{3 x ^{3}}) - \frac{1}{5 x ^{5}} - arctan (x)

Simplify

= - \frac{1}{x} + \frac{1}{3 x ^{3}} - \frac{1}{5 x ^{5}} - arctan (x)

Add a constant to the solution

= - \frac{1}{x} + \frac{1}{3 x ^{3}} - \frac{1}{5 x ^{5}} - arctan (x) + C

x \to 0 + lim (x (ln (4 x))^{2})

y = x^{4} + 5 x^{2} - x, (1, 5)

t d t + y e^{- t} d y = 0, y (0) = 1

y^{^{''}} + 4 y^{^{'}} + 3 y = e^{x}

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

What is the integral of 1/(x^8+x^6) ?
The integral of 1/(x^8+x^6) is -1/x+1/(3x^3)-1/(5x^5)-arctan(x)+C