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Q.
If $\int \frac{\left(8 x+\frac{8}{x}+9\right)}{\left(2 x^3+3 x^2+4 x\right)^2} d x=g(x)+C$, then the value of $9(g(1)-g(-1))$ is equal to
Integrals
Solution:
$\int \frac{\left(8 x+\frac{8}{x}+9\right) x^2}{\left(2 x^4+3 x^3+4 x^2\right)^2} d x=\int \frac{\left(8 x^3+9 x^2+8 x\right)}{\left(2 x^4+3 x^3+4 x^2\right)^2} d x$
Put $2 x ^4+3 x ^3+4 x ^2= t \Rightarrow dt =\left(8 x ^3+9 x ^2+8 x \right) dx$
$\int \frac{ dt }{ t ^2}=\frac{-1}{ t }+ C =\frac{-1}{2 x ^4+3 x ^3+4 x ^2}+ C$
$g (1)=\frac{-1}{9}, g (-1)=\frac{-1}{3} $
$g (1)- g (-1)=\frac{-1}{9}+\frac{1}{3}=\frac{2}{9} $
$9( g (1)- g (-1))=2$