Example 23 - Chapter 7 Class 12 Integrals

Transcript

Example 23 Find ﷐﷮﷮﷐﷮﷐𝑥﷮2﷯+2𝑥+5﷯﷯ 𝑑𝑥 ﷐﷮﷮﷐﷮﷐𝑥﷮2﷯+2𝑥+5﷯﷯ 𝑑𝑥 = ﷐﷮﷮﷐﷮﷐𝑥﷮2﷯+2𝑥+1+4﷯﷯ 𝑑𝑥 = ﷐﷮﷮﷐﷮﷐(𝑥+1)﷮2﷯+4﷯﷯ 𝑑𝑥 Let y = x + 1 ﷐𝑑𝑦 ﷮𝑑𝑥﷯=1 dy = dx Substituting = ﷐﷮﷮﷐﷮﷐𝑦﷮2﷯+4﷯﷯ 𝑑𝑦 = ﷐﷮﷮﷐﷮﷐𝑦﷮2﷯+﷐(2)﷮2﷯﷯﷯ 𝑑𝑦 = ﷐𝑦﷮2﷯ ﷐﷮﷐𝑦﷮2﷯+4﷯+ ﷐4﷮2﷯﷐ log﷮﷐𝑦+﷐﷮﷐𝑦﷮2﷯+4﷯﷯﷯+ C = ﷐𝑦﷮2﷯﷐﷮﷐𝑦﷮2﷯+4﷯+2 𝑙𝑜𝑔﷐𝑦+﷐﷮﷐𝑦﷮2﷯+4﷯﷯+ C Putting value of y = x + 1 = ﷐𝑥 + 1﷮2﷯ ﷐﷮﷐(𝑥+1)﷮2﷯ + 4﷯+2 𝑙𝑜𝑔﷐𝑥+1+﷐﷮﷐(𝑥+1)﷮2﷯ + 4﷯﷯+ C = ﷐𝑥 + 1﷮2﷯﷐﷮﷐𝑥﷮2﷯+1+2𝑥+ 4﷯+2 𝑙𝑜𝑔﷐𝑥+1+﷐﷮﷐𝑥﷮2﷯+1+2𝑥+ 4﷯﷯+ C = ﷐𝟏﷮𝟐﷯ (x + 1) ﷐﷮﷐𝒙﷮𝟐﷯ 𝟐𝒙+𝟓﷯+𝟐﷐𝐥𝐨𝐠﷮﷐𝒙+𝟏+﷐﷮﷐𝒙﷮𝟐﷯+𝟐𝒙+𝟓﷯﷯﷯+𝐂