Chapter 7 Class 12 Integrals
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Ex 7.2, 36 - Integrate (x + 1) (x + log x)^2 / x - Teachoo

Ex 7.2, 36 - Chapter 7 Class 12 Integrals - Part 2

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Ex 7.2, 36 Integrate ((๐‘ฅ + 1) (๐‘ฅ + logโก๐‘ฅ )^2)/๐‘ฅ โˆซ1โ–’ใ€–" " ((๐‘ฅ + 1) (๐‘ฅ + logโก๐‘ฅ )^2)/๐‘ฅใ€— . ๐‘‘๐‘ฅ Let ๐‘ฅ+logโก๐‘ฅ= ๐‘ก Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ 1+1/๐‘ฅ= ๐‘‘๐‘ก/๐‘‘๐‘ฅ (๐‘ฅ + 1)/๐‘ฅ= ๐‘‘๐‘ก/๐‘‘๐‘ฅ " " ๐‘‘๐‘ฅ = ((๐‘ฅ )/(๐‘ฅ + 1))๐‘‘๐‘ก Now, our function becomes โˆซ1โ–’ใ€–" " ((๐‘ฅ + 1) (๐‘ฅ + logโก๐‘ฅ )^2)/๐‘ฅใ€— . ๐‘‘๐‘ฅ Putting the value of ๐‘ฅโˆ’๐‘™๐‘œ๐‘”โก๐‘ฅ=๐‘ก & ๐‘‘๐‘ฅ=((๐‘ฅ )/(๐‘ฅ + 1))๐‘‘๐‘ก = โˆซ1โ–’ใ€–" " ((๐‘ฅ + 1) (๐‘ก)^2)/๐‘ฅใ€— . (๐‘ฅ )/((๐‘ฅ + 1) ) . ๐‘‘๐‘ก = โˆซ1โ–’ใ€–" " ๐‘ก^2 ใ€—. ๐‘‘๐‘ก" " = ๐‘ก^(2 + 1)/(2 + 1) +๐ถ = ๐‘ก^3/3 +๐ถ = ๐Ÿ/๐Ÿ‘ (๐’™+๐’๐’๐’ˆโก๐’™ )^๐Ÿ‘+๐‘ช (Using โˆซ1โ–’๐‘ฅ^๐‘› . ๐‘‘๐‘ฅ=๐‘ฅ^(๐‘›+1)/(๐‘› +1)) (Using ๐‘ก=๐‘ฅ+๐‘™๐‘œ๐‘”โก๐‘ฅ)

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