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Question Number 224600 by klipto last updated on 21/Sep/25
i am back guys
$$\mathrm{i}\:\mathrm{am}\:\mathrm{back}\:\mathrm{guys} \\ $$
Commented by fantastic last updated on 21/Sep/25
good
$${good} \\ $$
Commented by Tawa11 last updated on 21/Sep/25
Welcome back at Mathsdave.  Am I right sir??
$$\mathrm{Welcome}\:\mathrm{back}\:\mathrm{at}\:\mathrm{Mathsdave}. \\ $$$$\mathrm{Am}\:\mathrm{I}\:\mathrm{right}\:\mathrm{sir}?? \\ $$
Commented by klipto last updated on 21/Sep/25
thanks tawa osheun
$$\mathrm{thanks}\:\mathrm{tawa}\:\mathrm{osheun} \\ $$
Commented by Tawa11 last updated on 21/Sep/25
Yes sir.
$$\mathrm{Yes}\:\mathrm{sir}. \\ $$
Commented by Tawa11 last updated on 21/Sep/25
∫_( 0) ^( 1)  ((ln(x) ln(1  −  x^2 ) ln(1  +  x^2 ))/(1   −   x^2 )) dx  Please help sir.  You came at the right time.  Hahahahagaha.
$$\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{x}\right)\:\mathrm{ln}\left(\mathrm{1}\:\:−\:\:\mathrm{x}^{\mathrm{2}} \right)\:\mathrm{ln}\left(\mathrm{1}\:\:+\:\:\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{1}\:\:\:−\:\:\:\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx} \\ $$$$\mathrm{Please}\:\mathrm{help}\:\mathrm{sir}. \\ $$$$\mathrm{You}\:\mathrm{came}\:\mathrm{at}\:\mathrm{the}\:\mathrm{right}\:\mathrm{time}. \\ $$$$\mathrm{Hahahahagaha}. \\ $$

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