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Question Number 94530 by i jagooll last updated on 19/May/20

Commented by PRITHWISH SEN 2 last updated on 19/May/20
$∫((x^2 dx)/(x^3 .((x^3 +1))^(1/3) )) put x^3 +1 = t^3 ⇒x^2 dx=t^2 dt ∫(t/((t^3 −1)))dt=∫{(1/(3(t−1)))−((2t+1)/(6(t^2 +t+1)))+(1/(2(t^2 +t+1)))}dt =(1/3)ln∣t−1∣−(1/6)ln∣t^2 +t+1∣+(1/(√3))tan^(−1) ((2t+1)/(√3)) +C =(1/3)ln∣(x^3 +1)^(1/3) −1∣−(1/6)ln∣(x^3 +1)^(2/3) +(x^3 +1)^(1/3) +1∣+(1/(√3))tan^(−1) ((2(x^3 +1)^(1/3) +1)/(√3)) + C$
$\int \frac{x^{2} dx}{x^{3} . \sqrt[3]{x^{3} + 1}} put x^{3} + 1 = t^{3} \Rightarrow x^{2} dx = t^{2} dt$ $\int \frac{t}{(t^{3} - 1)} dt = \int {\frac{1}{3 (t - 1)} - \frac{2 t + 1}{6 (t^{2} + t + 1)} + \frac{1}{2 (t^{2} + t + 1)}} dt$ $= \frac{1}{3} \ln ∣ t - 1 ∣ - \frac{1}{6} \ln ∣ t^{2} + t + 1 ∣ + \frac{1}{\sqrt{3}} \tan^{- 1} \frac{2 t + 1}{\sqrt{3}} + C$ $= \frac{1}{3} \ln ∣ {(x^{3} + 1)}^{\frac{1}{3}} - 1 ∣ - \frac{1}{6} \ln ∣ {(x^{3} + 1)}^{\frac{2}{3}} + {(x^{3} + 1)}^{\frac{1}{3}} + 1 ∣ + \frac{1}{\sqrt{3}} \tan^{- 1} \frac{2 {(x^{3} + 1)}^{\frac{1}{3}} + 1}{\sqrt{3}} + C$
	Answered by MJS last updated on 19/May/20

	$\int \frac{d x}{x \sqrt[3]{x^{3} + 1}} =$ $[t = \frac{1}{\sqrt[3]{x^{3} + 1}} \to d x = - \frac{\sqrt[3]{{(x^{3} + 1)}^{4}}}{x^{2}}]$ $= \int \frac{d t}{t^{3} - 1} = \frac{1}{3} \int \frac{d t}{t - 1} - \frac{1}{3} \int \frac{t + 2}{t^{2} + t + 1} d t =$ $= \frac{1}{3} \int \frac{d t}{t - 1} - \frac{1}{6} \int \frac{2 t + 1}{t^{2} + t + 1} d t - \frac{1}{2} \int \frac{d t}{t^{2} + t + 1}$ $now it should be easy$
	Commented by john santu last updated on 19/May/20
	integral lover prof ��
	Commented by i jagooll last updated on 19/May/20

	$thank you prof$
	Answered by niroj last updated on 19/May/20
	$I= ∫ (dx/(x((x^3 +1))^(1/3) )) = ∫ (( 1)/(x(x^3 +1)^(1/3) ))dx Put, (x^3 +1)^(1/3) = t x^3 +1 =t^3 3x^2 dx=3t^2 dt dx=(t^2 /x^2 )dt⇒ dx= (t^2 /((t^3 −1)^(2/3) ))dt x^3 =t^3 −1 ⇒ x = (t^3 −1)^(1/3) I = ∫ (1/((t^3 −1)^(1/3) )). (t^2 /(t(t^3 −1)^(2/3) ))dt = ∫ (t/((t^3 −1)))dt = ∫ (((t−1)+1)/((t−1)(t^2 +t+1)))dt = ∫ (1/((t−1)(t^2 +t+1)))dt+∫ (1/(t^2 +t+1))dt = (1/3)∫ (1/(t−1))dt −(1/3)∫ ((t+2)/(t^2 +t+1))dt+∫ (1/(t^2 +t+1))dt = (1/3)∫(1/(t−1))dt −(1/6)∫ (((2t+1))/(t^2 +t+1))dt−(1/2)∫(1/(t^2 +t+1))dt+∫(1/(t^2 +t+1))dt = (1/3)∫ (1/(t−1))dt−(1/6)∫ ((2t+1)/(t^2 +t+1))dt+ (1/2)∫(1/(t^2 +t+1))dt = (1/3)log (t−1) −(1/6)log (t^2 +t+1)+(1/(√3))tan^(−1) ((2t+1)/(√3))+C Note: Put t= ((x^3 +1))^(1/3) = (1/3)log (((x^3 +1))^(1/3) −1)−(1/6)log {(((x^3 +1))^(1/3) )^2 +((x^3 +1))^(1/3) +1}+(1/(√3))tan^(−1) ((2 ((x^3 +1))^(1/3) +1)/(√3))+C//. Note:You can many changement in to line of simplify and their value of partial fraction to easier integrate if needed . it is just sort.$
	$I = \int \frac{dx}{x \sqrt[3]{x^{3} + 1}}$ $= \int \frac{1}{x {(x^{3} + 1)}^{\frac{1}{3}}} dx$ $Put, {(x^{3} + 1)}^{\frac{1}{3}} = t$ $x^{3} + 1 = t^{3}$ ${3 x}^{2} dx = {3 t}^{2} dt$ $dx = \frac{t^{2}}{x^{2}} dt \Rightarrow dx = \frac{t^{2}}{{(t^{3} - 1)}^{\frac{2}{3}}} dt$ $x^{3} = t^{3} - 1 \Rightarrow x = {(t^{3} - 1)}^{\frac{1}{3}}$ $I = \int \frac{1}{{(t^{3} - 1)}^{\frac{1}{3}}} . \frac{t^{2}}{t {(t^{3} - 1)}^{\frac{2}{3}}} dt$ $= \int \frac{t}{(t^{3} - 1)} dt$ $= \int \frac{(t - 1) + 1}{(t - 1) (t^{2} + t + 1)} dt$ $= \int \frac{1}{(t - 1) (t^{2} + t + 1)} dt + \int \frac{1}{t^{2} + t + 1} dt$ $= \frac{1}{3} \int \frac{1}{t - 1} dt - \frac{1}{3} \int \frac{t + 2}{t^{2} + t + 1} dt + \int \frac{1}{t^{2} + t + 1} dt$ $= \frac{1}{3} \int \frac{1}{t - 1} dt - \frac{1}{6} \int \frac{(2 t + 1)}{t^{2} + t + 1} dt - \frac{1}{2} \int \frac{1}{t^{2} + t + 1} dt + \int \frac{1}{t^{2} + t + 1} dt$ $= \frac{1}{3} \int \frac{1}{t - 1} dt - \frac{1}{6} \int \frac{2 t + 1}{t^{2} + t + 1} dt + \frac{1}{2} \int \frac{1}{t^{2} + t + 1} dt$ $= \frac{1}{3} \log (t - 1) - \frac{1}{6} \log (t^{2} + t + 1) + \frac{1}{\sqrt{3}} \tan^{- 1} \frac{2 t + 1}{\sqrt{3}} + C$ $Note : Put t = \sqrt[3]{x^{3} + 1}$ $= \frac{1}{3} \log (\sqrt[3]{x^{3} + 1} - 1) - \frac{1}{6} \log {{(\sqrt[3]{x^{3} + 1})}^{2} + \sqrt[3]{x^{3} + 1} + 1} + \frac{1}{\sqrt{3}} \tan^{- 1} \frac{2 \sqrt[3]{x^{3} + 1} + 1}{\sqrt{3}} + C / / .$ $N o t e : Y o u c a n m a n y c h a n g e m e n t i n t o l i n e o f s i m p l i f y a n d t h e i r v a l u e o f p a r t i a l f r a c t i o n t o e a s i e r i n t e g r a t e i f n e e d e d . i t i s j u s t s o r t .$
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