Q83 / 160 · TS EAMCET 2010 · Mathematics

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

  • Correct answer
    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
    $\frac{-1}{1+x+x^8}$

Solution

$\int \frac{7 x^8+8 x^7}{\left(1+x+x^8\right)^2} d x=f(x)+c$ 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, $f(x)=\frac{d}{d x}\left\{\frac{x^8}{\left(1+x+x^8\right)}\right\}$ $=\frac{\left[\left(1+x+x^8\right) \cdot 8 x^7-x^8\left(1+8 x^7\right)\right]}{\left(1+x+x^8\right)^2}$ $f(x)=\frac{\left(8 x^7+8 x^8+8 x^{15}-x^8-8 x^{15}\right)}{\left(1+x+x^8\right)^2}$ $f(x)=\frac{\left(7 x^8+8 x^7\right)}{\left(1+x+x^8\right)^2}$ Hence, $\int \frac{7 x^8+8 x^7}{\left(1+x+x^8\right)^2} d x=\frac{x^8}{\left(1+x+x^8\right)}+c$ OBJECT END]

Hence, the correct answer is option: 'A'.

Relevant Questions