∫(dx/(x(x^(2018) +1)))


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Question Number 35456 by math1967 last updated on 19/May/18

$\int \frac{d x}{x (x^{2018} + 1)}$
	Answered by ajfour last updated on 19/May/18

	$I = - \frac{1}{2018} \int \frac{- 2018 x^{- 19} d x}{1 + x^{- 2018}}$ $= - \frac{1}{2018} \ln ∣ 1 + \frac{1}{x^{2018}} ∣ + c .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 19/May/18
	$multiply N_(r ) and D_(r ) by x^(2017) ∫((x^(2017 ) dx)/(x^(2018) (1+x^(2018) ))) t=x^(2018 ) so dt=2018×x^(2017) dx (1/(2018))∫(dt/(t(t+1))) =(1/(2018))∫(((t+1)−t)/(t(t+1))) dt =(1/(2018)){∫(dt/t)−∫(dt/(t+1))} =(1/(2018)){lnt−ln(t+1)} =(1/(2018))ln((t/(t+1))) =(1/(2018))ln((x^(2018) /(x^(2018) +1)))$
	$m u l t i p l y N_{r} a n d D_{r} b y x^{2017}$ $\int \frac{x^{2017} d x}{x^{2018} (1 + x^{2018})}$ $t = x^{2018} s o d t = 2018 \times x^{2017} d x$ $\frac{1}{2018} \int \frac{d t}{t (t + 1)}$ $= \frac{1}{2018} \int \frac{(t + 1) - t}{t (t + 1)} d t$ $= \frac{1}{2018} {\int \frac{d t}{t} - \int \frac{d t}{t + 1}}$ $= \frac{1}{2018} {l n t - l n (t + 1)}$ $= \frac{1}{2018} l n (\frac{t}{t + 1})$ $= \frac{1}{2018} l n (\frac{x^{2018}}{x^{2018} + 1})$
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