Misc 4 - Chapter 7 Class 12 Integrals

Transcript

Misc 4 Integrate the function 1﷮ 𝑥﷮2﷯ 𝑥﷮4﷯ + 1﷯﷮ 3﷮4﷯﷯ ﷯ ﷮﷮ 1 𝑑𝑥﷮ 𝑥﷮2﷯ 𝑥﷮4﷯ + 1﷯﷮ 3﷮4﷯﷯ ﷯﷯ Taking 𝒙﷮𝟒﷯ common from denominator = ﷮﷮ 1 𝑑𝑥﷮ 𝑥﷮2 ﷯ (𝑥﷮4﷯)﷮ 3﷮4﷯﷯ 1 + 1﷮ 𝑥﷮4﷯﷯﷯﷮ 3﷮4﷯﷯﷯﷯ = ﷮﷮ 𝑑𝑥﷮ 𝑥﷮2﷯ (𝑥﷮3﷯) 1 + 1﷮ 𝑥﷮4﷯﷯﷯﷮ 3﷮4﷯﷯﷯﷯ = ﷮﷮ 𝑑𝑥﷮ 𝑥﷮5﷯ 1 + 1﷮ 𝑥﷮4﷯﷯﷯﷮ 3﷮4﷯﷯﷯﷯ Let t = 1 + 1﷮ 𝑥﷮4﷯﷯ 𝑑𝑡﷮𝑑𝑥﷯ = − 4﷮ 𝑥﷮5﷯﷯ − 𝑑𝑡﷮4﷯ = 𝑑𝑥﷮ 𝑥﷮5﷯﷯ Substituting value of x and dx = −1﷮4﷯ ﷮﷮ 𝑑𝑡﷮ 𝑡﷮ 3﷮4﷯﷯﷯﷯ = −1﷮4﷯ ﷮﷮ 𝑡﷮ −3﷮4﷯﷯﷯ 𝑑𝑡 = −1﷮4﷯ 𝑡﷮ −3﷮4﷯ + 1﷯﷮ −3﷮4﷯ + 1﷯﷯+ C = −1﷮4﷯ 𝑡﷮ 1﷮4﷯﷯﷮ 1﷮4﷯﷯﷯+ C = − 𝑡﷮ 1﷮4﷯﷯+ C = − 𝟏+ 𝟏﷮ 𝒙﷮𝟒﷯﷯﷯﷮ 𝟏﷮𝟒﷯﷯+ C