Miscellaneous

Chapter 7 Class 12 Integrals
Serial order wise

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Misc 5 Integrate the function 1/( ๐ฅ^(1/2) + ๐ฅ^(1/3) ) โซ1โ1/(๐ฅ^(1/2) + ๐ฅ^(1/3) ) ๐๐ฅ Let x = ๐ก^6 ๐๐ฅ/๐๐ก=6๐ก^5 dx = 6๐ก^5 dt Substituting value of x and dx the equation โซ1โ1/(๐ฅ^(1/2) + ๐ฅ^(1/3) ) ๐๐ฅ = โซ1โใ6๐กใ^5/(ใใ(๐กใ^6) ใ^(1/2) +ใใ (๐กใ^6) ใ^(1/3) ) = โซ1โใ6๐กใ^5/(๐ก^3 + ๐ก^2 ) ๐๐ก = 6โซ1โ๐ก^3/(๐ก + 1) ๐๐ก Adding and subtracting 1 in numerator = 6โซ1โ(๐ก^3 + 1 โ 1)/(๐ก + 1) ๐๐ก = 6โซ1โ((๐ก^3 + 1 )/(๐ก + 1)โ1/(๐ก + 1)) ๐๐ก Using ๐^3+๐^3=(๐+๐)(๐^2+๐^2โ๐๐) = 6โซ1โ((๐ก + 1)(๐ก^2+1^2 โ 1 ร ๐ก))/(๐ก + 1)โ1/(๐ก + 1) ๐๐ก = 6โซ1โใ(๐ก^2+1โ๐ก)โ1/(1 + ๐ก)ใ ๐๐ก = 6[๐ก^3/3+๐กโ๐ก^2/2โ๐๐๐|1+๐ก|]+ C = 2๐ก^3+6๐กโใ3๐กใ^2โ6 ๐๐๐|1+๐ก|+ C Putting back value of t = 2 ใใ(๐ฅใ^(1/6))ใ^3+6ใ(๐ฅใ^(1/6))โ3 ใใ(๐ฅใ^(1/6))ใ^2โ6 ๐๐๐|1+๐ฅ^(1/6) |+ C = 2๐ฅ^(1/2)+6๐ฅ^(1/6)โ3๐ฅ^(1/3)โ6 ๐๐๐|1+๐ฅ^(1/6) |+ C = 2โ๐โ๐๐^(๐/๐)+ใ๐ ๐ใ^(๐/๐)โ๐ ๐ฅ๐จ๐ โกใ(๐+๐^(๐/๐))ใ+ C Because (1+ใ๐ฅ ใ^(1/6)) is always positive