Find : โˆซโˆš(x/(1-x^3 )) dx; xโˆˆ(0,1)

This question is similar to Misc 12 Chapter 7 Class 12

[Class 12] Find โˆซ โˆš(x/1-x^3) dx - Teachoo Sample Paper (with Video) - CBSE Class 12 Sample Paper for 2024 Boards

part 2 - Question 28 (Choice 1) - CBSE Class 12 Sample Paper for 2024 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12

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โˆซโˆš(๐‘ฅ/(1 โˆ’ ๐‘ฅ^3 )) ๐‘‘๐‘ฅ = โˆซ1โ–’โˆš๐‘ฅ/โˆš(1 โˆ’ ๐‘ฅ^3 ) ๐‘‘๐‘ฅ Let ๐’•=๐’™^(๐Ÿ‘/๐Ÿ) Differentiating w.r.t. ๐‘ฅ ๐‘‘๐‘ก/๐‘‘๐‘ฅ=3/2 ๐‘ฅ^(3/2 โˆ’1) ๐‘‘๐‘ก/๐‘‘๐‘ฅ=3/2 ๐‘ฅ^(1/2 ) ๐‘‘๐‘ก/๐‘‘๐‘ฅ=3/2 โˆš๐‘ฅ ๐Ÿ/๐Ÿ‘ ๐๐ญ=โˆš๐’™ ๐’…๐’™ Now, our equation becomes โˆซ1โ–’โˆš๐‘ฅ/โˆš(1 โˆ’ ๐‘ฅ^3 ) ๐‘‘๐‘ฅ=โˆซ1โ–’ใ€–2/(3โˆš(1 โˆ’ใ€– ๐‘ฅใ€—^3 )) ๐‘‘๐‘กใ€— = ๐Ÿ/๐Ÿ‘ โˆซ1โ–’๐’…๐’•/โˆš(๐Ÿ โˆ’ ๐’•^๐Ÿ )dt = 2/3 sin^(โˆ’1)โกใ€–๐‘ก+๐‘ใ€— Putting back ๐‘ก=๐‘ฅ^(3/2) = ๐Ÿ/๐Ÿ‘ ใ€–๐’”๐’Š๐’ใ€—^(โˆ’๐Ÿ)โกใ€–ใ€–(๐’™ใ€—^(๐Ÿ‘/๐Ÿ))+๐’„ใ€—

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 15 years. He provides courses for Maths, Science and Computer Science at Teachoo