Question

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

• A. $\frac{x^8}{1+x+x^8}$
• B. $28\log (1+x+x^8)$
• C. $\frac{1}{1+x+x^8}$
• D. $\frac{-1}{1+x+x^8}$

Answer 1

Answer: (A)

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

We have

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

$\therefore \int$ $\frac{8x^7+7x^8}{(1+x+x^8{)}^2}$ $dx=$ $\int$ $d\left(\frac{x^8}{1+x+x^8}\right)=$ $\left(\frac{x^8}{1+x+x^8}\right)+c$.

Comparing it with the given problem we have

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