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Popular >

integral of x/(16-x^8)

Solution

\int \frac{x}{16 - x ^{8}} d x

Solution

- \frac{1}{32} arctan (x + 1) + \frac{1}{32} arctan (x - 1) + \frac{1}{64} ln x^{2} + 2 - \frac{1}{64} ln x + 2 - \frac{1}{64} ln x - 2 + C

Solution steps

\int \frac{x}{16 - x ^{8}} d x

Take the partial fraction of

\frac{x}{16 - x ^{8}} : - \frac{1}{32 ( x ^{2} + 2 x + 2 )} + \frac{1}{32 ( x ^{2} - 2 x + 2 )} + \frac{x}{32 ( x ^{2} + 2 )} - \frac{1}{64 ( x + 2 )} - \frac{1}{64 ( x - 2 )}

= \int - \frac{1}{32 ( x ^{2} + 2 x + 2 )} + \frac{1}{32 ( x ^{2} - 2 x + 2 )} + \frac{x}{32 ( x ^{2} + 2 )} - \frac{1}{64 ( x + 2 )} - \frac{1}{64 ( x - 2 )} 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}{32 ( x ^{2} + 2 x + 2 )} d x + \int \frac{1}{32 ( x ^{2} - 2 x + 2 )} d x + \int \frac{x}{32 ( x ^{2} + 2 )} d x - \int \frac{1}{64 ( x + 2 )} d x - \int \frac{1}{64 ( x - 2 )} d x

\int \frac{1}{32 ( x ^{2} + 2 x + 2 )} d x = \frac{1}{32} arctan (x + 1)

\int \frac{1}{32 ( x ^{2} - 2 x + 2 )} d x = \frac{1}{32} arctan (x - 1)

\int \frac{x}{32 ( x ^{2} + 2 )} d x = \frac{1}{64} ln x^{2} + 2

\int \frac{1}{64 ( x + 2 )} d x = \frac{1}{64} ln x + 2

\int \frac{1}{64 ( x - 2 )} d x = \frac{1}{64} ln x - 2

= - \frac{1}{32} arctan (x + 1) + \frac{1}{32} arctan (x - 1) + \frac{1}{64} ln x^{2} + 2 - \frac{1}{64} ln x + 2 - \frac{1}{64} ln x - 2

Add a constant to the solution

= - \frac{1}{32} arctan (x + 1) + \frac{1}{32} arctan (x - 1) + \frac{1}{64} ln x^{2} + 2 - \frac{1}{64} ln x + 2 - \frac{1}{64} ln x - 2 + C

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Frequently Asked Questions (FAQ)

What is the integral of x/(16-x^8) ?
The integral of x/(16-x^8) is