∫7x8+8x71+x+x82dx=fx+c⇒fx=

∫7x8+8x71+x+x8dx=fx+c=∫7x8+8x7x821x8+1x7+12dx=∫7x8+8x9dx1x8+1x7+12, 1x8+1x7+1=t=∫−dtt2=1t+cfx=1t=x81+x+x8

∫7x8+8x71+x+x82dx=fx+c⇒fx=

∫7x8+8x71+x+x8dx=fx+c=∫7x8+8x7x821x8+1x7+12dx=∫7x8+8x9dx1x8+1x7+12, 1x8+1x7+1=t=∫−dtt2=1t+cfx=1t=x81+x+x8

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$\int \frac{7 x^{8} + 8 x^{7}}{{(1 + x + x^{8})}^{2}} d x = f (x) + c \Rightarrow f (x) =$

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$\frac{x^{8}}{1 + x + x^{8}}$

$28 \log (1 + x + x^{8})$

$\frac{1}{1 + x + x^{8}}$

$- \frac{1}{1 + x + x^{8}}$

answer is A.

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Detailed Solution

$\int \frac{7 x^{8} + 8 x^{7}}{(1 + x + x^{8})} d x = f (x) + c$

$= \int \frac{7 x^{8} + 8 x^{7}}{{(x^{8})}^{2} {[\frac{1}{x^{8}} + \frac{1}{x^{7}} + 1]}^{2}} d x$

$= \int \frac{(\frac{7}{x^{8}} + \frac{8}{x^{9}}) d x}{{(\frac{1}{x^{8}} + \frac{1}{x^{7}} + 1)}^{2}}, \frac{1}{x^{8}} + \frac{1}{x^{7}} + 1 = t$

$= \int \frac{- d t}{t^{2}} = \frac{1}{t} + c$

$f (x) = \frac{1}{t} = \frac{x^{8}}{1 + x + x^{8}}$

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$\int \frac{7 x^{8} + 8 x^{7}}{{(1 + x + x^{8})}^{2}} d x = f (x) + c \Rightarrow f (x) =$