∫7x8+8x7(1+x+x8)2dx=f(x)+c⇒f(x)=

ddx{x81+x+x8}=(1+x+x8)(8.x7)−(x8)(8.x7+1)(1+x+x8)2=7x8+8x7(1+x+x8)2⇒∫7x8+8x7(1+x+x8)2dx=x81+x+x8+c∴f(x)=x81+x+x8

∫7x8+8x7(1+x+x8)2dx=f(x)+c⇒f(x)=

ddx{x81+x+x8}=(1+x+x8)(8.x7)−(x8)(8.x7+1)(1+x+x8)2=7x8+8x7(1+x+x8)2⇒∫7x8+8x7(1+x+x8)2dx=x81+x+x8+c∴f(x)=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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$\frac{- 1}{1 + x + x^{8}}$

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

$28. (1 + x + x^{8})$

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

answer is A.

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$\frac{d}{d x} {\frac{x^{8}}{1 + x + x^{8}}} = \frac{(1 + x + x^{8}) (8. x^{7}) - (x^{8}) (8. x^{7} + 1)}{{(1 + x + x^{8})}^{2}}$

$= \frac{7 x^{8} + 8 x^{7}}{{(1 + x + x^{8})}^{2}}$

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

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

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