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Q.
If $\int \frac{7 x^{8}+8 x^{7}}{\left(1+x+x^{8}\right)^{2}} d x=f(x)+c_{2}$
then $f(x)$ is equal to
EAMCETEAMCET 2010
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, $\int \frac{7 x^{8}+ 8 x^{7}}{\left(1+x+x^{8}\right)^{2}} d x=\frac{x^{8}}{\left(1+x+x^{8}\right)}+c$