Ex 7.2

Chapter 7 Class 12 Integrals
Serial order wise

### Transcript

Ex 7.2, 7 Integrate the function: ๐ฅโ(๐ฅ+2) ๐ฅโ(๐ฅ+2) Step 1: Let (๐ฅ+2)=๐ก Differentiating both sides ๐ค.๐.๐ก.๐ฅ 1+0 = ๐๐ก/๐๐ฅ 1= ๐๐ก/๐๐ฅ ๐๐ฅ=๐๐ก Step 2: Integrating the function โซ1โใ" " ๐ฅโ(๐ฅ+2)ใ .๐๐ฅ Putting the value of ๐ฅ+2 & ๐๐ฅ . = โซ1โใ๐ฅโ๐กใ .๐๐ฅ = โซ1โใ๐ฅโ๐กใ .๐๐ก = โซ1โใ(๐กโ2) โ๐กใ .๐๐ก = โซ1โใ(๐กโ2) ๐ก^(1/2) ใ .๐๐ก = โซ1โ(๐ก.๐ก^(1/2)โ2.๐ก^(1/2) ) .๐๐ก = โซ1โ(๐ก^(3/2)โ2.๐ก^(1/2) ) .๐๐ก = โซ1โ๐ก^(3/2) .๐๐ก โ 2โซ1โ๐ก^(1/2) .๐๐ก (Using ๐ฅ+2=๐ก, ๐ฅ=๐กโ2) = ๐ก^(3/2 + 1)/(3/2 + 1) โ 2 . ๐ก^(1/2 + 1)/(1/2 + 1) + ๐ถ = ๐ก^(5/2)/(5/2) โ 2 . ๐ก^(3/2)/(3/2) + ๐ถ = 2/5 ๐ก^(5/2) โ 2 ร 2/3 ๐ก^(3/2) + ๐ถ = 2/5 ๐ก^(5/2) โ 4/3 ๐ก^(3/2) + ๐ถ Putting back ๐ก=๐ฅ+2 = ๐/๐ (๐+๐)^(๐/๐) โ ๐/๐ (๐+๐)^(๐/๐) + ๐ช (Using โซ1โ๐ฅ^๐ . ๐๐ฅ=๐ฅ^(๐+1)/(๐ +1) )