If ∫ (7x¹⁰ + 9x⁸) / (1 + x² + 2x⁹)² dx = f(x) + c and f(1) = 1/4, then f(x) is

The correct option is:
x⁹ / (1 + x² + 2x⁹)

Step-by-step Derivation:

We are given the indefinite integral:
$I=\int \frac{7x^{10}+9x^8}{{\left(1+x^2+2x^9\right)}^2}$

To evaluate this integral, we can simplify the expression inside the denominator. Let us factor out $x^9$ from the term inside the square in the denominator:
$1+x^2+2x^9=x^9\left(x^{-9}+x^{-7}+2\right)$

Squaring this term gives:
${\left(1+x^2+2x^9\right)}^2=x^{18}{\left(x^{-9}+x^{-7}+2\right)}^2$

Now, substitute this back into the integral:
$I=\int \frac{7x^{10}+9x^8}{x^{18}{\left(x^{-9}+x^{-7}+2\right)}^2}dx$

Divide each term in the numerator by $x^{18}$:
$I=\int \frac{7x^{-8}+9x^{-10}}{{\left(x^{-9}+x^{-7}+2\right)}^2}$

Now, we can use the method of substitution. Let:
$u=x^{-9}+x^{-7}+2$

Differentiating both sides with respect to $x$:
$du=\left(-9x^{-10}-7x^{-8}\right)dx$
$du=-\left(7x^{-8}+9x^{-10}\right)dx$
$-du=\left(7x^{-8}+9x^{-10}\right)dx$

Substituting these into the integral gives:
$I=\int \frac{-du}{u^2}=\int -u^{-2}du$

Integrating with respect to $u$:
$I=\frac{1}{u}+c$

Substituting back $u=x^{-9}+x^{-7}+2$:
$I=\frac{1}{x^{-9}+x^{-7}+2}+c$

To simplify the fraction, multiply the numerator and denominator by $x^9$:
$I=\frac{x^9}{1+x^2+2x^9}+c$

Therefore, comparing this to the given form $I=f\left(x\right)+c$, we have:
$f\left(x\right)=\frac{x^9}{1+x^2+2x^9}$

We verify the condition $f\left(1\right)=\frac{1}{4}$:
$f\left(1\right)=\frac{1^9}{1+1^2+2{\left(1\right)}^9}=\frac{1}{1+1+2}=\frac{1}{4}$

The condition holds true. Thus, the function is:
$f\left(x\right)=\frac{x^9}{1+x^2+2x^9}$