Question

If $\int \frac{7x^8+8x^7}{{\left(1+x+x^8\right)}^2}dx=f(x)+c_2$ then $f(x)$ is equal to

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

Answer 1

Answer: (A)

$\int \frac{7x^8+8x^7}{{\left(1+x+x^8\right)}^2}dx=f(x)+c$
here better way to solve this integration we will check the options ie,
we will differentiate the option one by one and check the integration function
Taking option (a) we see that,
$f(x)=\frac{d}{dx}\left\{\frac{x^8}{\left(1+x+x^8\right)}\right\}$
$=\frac{\left[\left(1+x+x^8\right)\cdot 8x^7-x^8\left(1+8x^7\right)\right]}{{\left(1+x+x^8\right)}^2}$
$f(x)=\frac{\left(8x^7+8x^8+8x^{15}-x^8-8x^{15}\right)}{{\left(1+x+x^8\right)}^2}$
$f(x)=\frac{\left(7x^8+8x^7\right)}{{\left(1+x+x^8\right)}^2}$
Hence, $\int \frac{7x^8+8x^7}{{\left(1+x+x^8\right)}^2}dx=\frac{x^8}{\left(1+x+x^8\right)}+c$