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Question Number 43319 by Meritguide1234 last updated on 09/Sep/18

	Answered by MJS last updated on 09/Sep/18

	$\frac{8 x + \frac{8}{x} + 9}{{(2 x^{3} + 3 x^{2} + 4 x)}^{2}} = \frac{8 x^{2} + 9 x + 8}{x {(2 x^{3} + 3 x^{2} + 4 x)}^{2}} =$ $= \frac{8 x^{3} + 9 x^{2} + 8 x}{{(2 x^{4} + 3 x^{3} + 4 x^{2})}^{2}} = \frac{f^{'} (x)}{f {(x)}^{2}}$ $\int \frac{f^{'} (x)}{f {(x)}^{2}} d x = - f {(x)}^{- 1}$ $\int \frac{8 x + \frac{8}{x} + 9}{{(2 x^{3} + 3 x^{2} + 4 x)}^{2}} d x = - \frac{1}{2 x^{4} + 3 x^{3} + 4 x^{2}} + C$
	Commented by Meritguide1234 last updated on 10/Sep/18

	$t h a n k y o u$
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