Question 15

The value of $$\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}$$

Solution

$$\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}$$
or $$\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}$$
or $$\frac{4}{1-x^4}+\frac{4}{1+x^4}$$
or $$\frac{8}{1-x^8}$$

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The value of $$\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}$$

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