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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
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Verified Answer
The correct answer is:
$\frac{x^8}{1+x+x^8}$
$\int \frac{7 x^8+8 x^7}{\left(1+x+x^8\right)^2} d x=f(x)+c$
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]
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]
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