Ex 7.4, 17 - Chapter 7 Class 12 Integrals (Term 2)

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Ex 7.4, 17 Integrate the function (𝑥 + 2)/√(𝑥^2 − 1) ∫1▒(𝑥 + 2)/√(𝑥^2 − 1) . 𝑑𝑥=∫1▒(1/2 (2𝑥) + 2" " )/√(𝑥^2 − 1) . 𝑑𝑥 =∫1▒(1/2 (2𝑥))/√(𝑥^2 − 1) . 𝑑𝑥+∫1▒2/√(𝑥^2 − 1) . 𝑑𝑥 =1/2 ∫1▒( 2𝑥)/√(𝑥^2 − 1) . 𝑑𝑥+∫1▒2/√(𝑥^2 − 1) . 𝑑𝑥 Solving 𝑰𝟏 I1=1/2 ∫1▒( 2𝑥)/√(𝑥^2 − 1) 𝑑𝑥 …(1) Let 𝑥^2−1=𝑡 Differentiating w.r.t. x 2𝑥−0=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/2𝑥 Thus, our equation becomes I1=1/2 ∫1▒( 2𝑥)/√(𝑥^2 − 1) 𝑑𝑥 Put the values of (𝑥^2−1)=𝑡 and 𝑑𝑥, we get I1=1/2 ∫1▒( 2𝑥)/√𝑡 𝑑𝑥 I1=1/2 ∫1▒( 2𝑥)/√𝑡 × 𝑑𝑡/2𝑥 I1=1/2 ∫1▒( 1)/√𝑡 𝑑𝑡 I1=1/2 ∫1▒1/𝑡^(1/2) 𝑑𝑡 I1=1/2 ∫1▒𝑡^((−1)/2) 𝑑𝑡 I1=1/2 𝑡^((−1)/2 + 1)/((−1)/2 + 1) +𝐶1 I1=1/2 𝑡^(1/2)/(1/2) +𝐶1 I1=𝑡^(1/2)+𝐶1 I1=√𝑡+𝐶1 I1=√(𝑥^2 − 1) + 𝐶1 Solving 𝑰𝟐 I2=∫1▒2/√(𝑥^2 − 1) . 𝑑𝑥 I2=2∫1▒1/√(𝑥^2 − (1)^2 ) . 𝑑𝑥 I2=2 𝑙𝑜𝑔⁡|𝑥+√(𝑥^2 −1)|+𝐶2 It is of form ∫1▒𝑑𝑥/√(𝑥^2 − 𝑎^2 ) =𝑙𝑜𝑔⁡|𝑥+√(𝑥^2 − 𝑎^2 )|+𝐶 ∴ Replacing a by 1 , we get ("Using " 𝑡=𝑥^2−1) Now, Putting the values of I1 and I2 in (1) ∫1▒(𝑥 + 2)/√(𝑥^2 − 1) . 𝑑𝑥=1/2 ∫1▒( 2𝑥)/√(𝑥^2 − 1) . 𝑑𝑥+2∫1▒1/√(𝑥^2 − 1) . 𝑑𝑥 =√(𝑥^2 − 1) + 𝐶1+2 𝑙𝑜𝑔⁡|𝑥+√(𝑥^2 −1) |+𝐶2 =√(𝒙^𝟐 − 𝟏)+𝟐 𝒍𝒐𝒈⁡|𝒙+√(𝒙^𝟐 −𝟏)|+ 𝑪