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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Transcript

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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