1. Class 12
2. Important Question for exams Class 12

Transcript

Misc 24 Integrate the function ﷮ 𝑥﷮2﷯ + 1﷯ log﷮ 𝑥﷮2﷯+ 1﷯ − 2 log﷮𝑥﷯﷯﷯ ﷮ 𝑥﷮4﷯﷯ ﷮﷮ ﷮ 𝑥﷮2﷯ + 1﷯ log﷮ 𝑥﷮2﷯+ 1﷯ −2 log﷮𝑥﷯﷯﷯ ﷮ 𝑥﷮4﷯﷯﷯𝑑𝑥 Taking 𝑥﷮2﷯common from ﷮ 𝑥﷮2﷯+1﷯ = ﷮﷮ (𝑥﷮2﷯) ﷮ 1﷮2﷯﷯ 1 + 1﷮ 𝑥﷮2﷯﷯﷯﷮ 1﷮2﷯﷯ log﷮( 𝑥﷮2﷯+1)﷯ − log﷮ 𝑥﷮2﷯﷯﷯﷮ 𝑥﷮4﷯﷯﷯ 𝑑𝑥 = ﷮﷮ 𝑥 1+ 1﷮ 𝑥﷮2﷯﷯﷯﷮ 1﷮2﷯﷯ log﷮ 𝑥﷮2 ﷯+ 1﷯﷮ 𝑥﷮2﷯﷯﷯﷯﷮ 𝑥﷮4﷯﷯﷯ 𝑑𝑥 = ﷮﷮ 1+ 1﷮ 𝑥﷮2﷯﷯﷯﷮ 1﷮2﷯﷯ log﷮ 1 + 1﷮ 𝑥﷮2﷯﷯﷯﷯﷯﷮ 𝑥﷮3﷯﷯﷯ Let t = 1 + 1﷮ 𝑥﷮2﷯﷯ 𝑑𝑡﷮𝑑𝑥﷯= −2﷮ 𝑥﷮3﷯﷯ −1﷮2﷯𝑑𝑡= 𝑑𝑥﷮ 𝑥﷮3﷯﷯ Substituting, = − 1﷮2﷯ ﷮﷮ 𝑡﷮ 1﷮2﷯﷯﷯ log 𝑡﷮ 𝑑𝑡﷯ Hence, − 1﷮2﷯ ﷮﷮ 𝑡﷮ 1﷮2﷯﷯ log﷮𝑡 𝑑𝑡=− 1﷮2﷯ log﷮𝑡 ﷮﷮ 𝑡﷮ 1﷮2﷯﷯ 𝑑𝑡﷯− ﷮﷮ 𝑑( log﷮𝑡)﷯﷮𝑑𝑡﷯ ﷮﷮ 𝑡﷮ 1﷮2﷯﷯﷯ 𝑑𝑡﷯﷯ 𝑑𝑡﷯﷯﷯﷯ = − 1﷮2﷯ log﷮𝑡 2 𝑡﷮ 3﷮2﷯﷯﷮3﷯﷯− ﷮﷮ 1﷮𝑡﷯× 2﷮3﷯ 𝑡﷮ 3﷮2﷯﷯﷯﷯ 𝑑𝑡﷯ = − 1﷮2﷯ 2﷮3﷯ 𝑡﷮ 3﷮2﷯﷯ log﷮𝑡− 2﷮3﷯﷯ ﷮﷮ 𝑡﷮ 1﷮2﷯﷯ 𝑑𝑡﷯﷯ = − 1﷮2﷯ 2﷮3﷯ 𝑡﷮ 3﷮2﷯﷯ log﷮𝑡− 2﷮3﷯﷯ 2𝑡﷮ 3﷮2﷯﷯﷮3﷯﷯ = 2﷮9﷯ 𝑡﷮ 3﷮2﷯﷯− 1﷮3﷯ 𝑡﷮ 3﷮2﷯﷯ log﷮𝑡﷯ Putting value of t = 2﷮9﷯ 1+ 1﷮ 𝑥﷮2﷯﷯﷯﷮ 3﷮2﷯﷯− 1﷮3﷯ 1+ 1﷮ 𝑥﷮2﷯﷯﷯﷮ 3﷮2﷯﷯ log﷮ 1+ 1﷮ 𝑥﷮2﷯﷯﷯+﷯ C = − 𝟏﷮𝟑﷯ 𝟏+ 𝟏﷮ 𝒙﷮𝟐 ﷯﷯﷯﷮ 𝟑﷮𝟐﷯﷯ 𝐥𝐨𝐠﷮ 𝟏+ 𝟏﷮ 𝒙﷮𝟐﷯﷯﷯﷯− 𝟐﷮𝟑﷯﷯+ C

Class 12
Important Question for exams Class 12