Ex 7.2, 11 - Chapter 7 Class 12 Integrals

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Ex 7.2, 11 (Method 1) Integrate the function: 𝑥/√(𝑥 + 4) , 𝑥>0 Step 1: Simplify the given function 𝑥/√(𝑥 + 4) = (𝑥 + 4 − 4)/√(𝑥 + 4) = (𝑥 + 4)/√(𝑥 + 4) − 4/√(𝑥 + 4) = (𝑥 + 4)^(1/2 − 1) − 4(𝑥 + 4)^(1/2) = (𝑥 + 4)^(1/2) − 4(𝑥 + 4)^(− 1/2) (Adding and Subtracting 4) Step 2: Integrating the function ∫1▒〖" " 𝑥/√(𝑥 + 4)〗 . 𝑑𝑥 = ∫1▒((𝑥 + 4)^(1/2) " − " 〖4 (𝑥 + 4)〗^(− 1/2) ) . 𝑑𝑥 = ∫1▒(𝑥 + 4)^(1/2) . 𝑑𝑥 − 4∫1▒(𝑥 + 4)^(− 1/2) . 𝑑𝑥 = (𝑥 + 4)^(1/2 + 1)/(1/2 + 1) − (4 (𝑥 + 4)^(− 1/2 + 1))/(− 1/2 + 1) + C = (𝑥 + 4)^(3/2)/(3/2 ) − (4 (𝑥 + 4)^(1/2))/( 1/2) + C = 2/3 (𝑥+4)^(3/2) − 4.2 (𝑥+4)^(1/2) + C = 〖2(𝑥+4)〗^(1/2) ((𝑥 + 4)/3 −4) + 𝐶 = 〖2(𝑥+4)〗^(1/2) ((𝑥 + 4 −12)/3) + 𝐶 = 〖2(𝑥+4)〗^(1/2) ((𝑥 − 8))/3 + 𝐶 = 𝟐/𝟑 √(𝒙 + 𝟒) (𝒙−𝟖) + 𝑪 (Taking 〖2(𝑥+4)〗^(1/2) as common) Ex 7.2, 11 (Method 2) Integrate the function: 𝑥/√(𝑥 + 4) , 𝑥>0 Step 1: Let 𝑥+4=𝑡 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 1+0= 𝑑𝑡/𝑑𝑥 1= 𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡 Step 2: Integrating the function ∫1▒〖" " 𝑥/√(𝑥 + 4)〗 . 𝑑𝑥 Putting 𝑡=𝑥+4 & 𝑑𝑥=𝑑𝑡 = ∫1▒𝑥/√𝑡 . 𝑑𝑡 = ∫1▒(𝑡 − 4)/√𝑡 . 𝑑𝑡 = ∫1▒(𝑡 − 4)/√𝑡 . 𝑑𝑡 = ∫1▒(𝑡/√𝑡 −4/√𝑡) 𝑑𝑡 = ∫1▒𝑡/√𝑡 . 𝑑𝑡 − ∫1▒4/√𝑡 . 𝑑𝑡 = ∫1▒𝑡^(1 − 1/2) . 𝑑𝑡 − ∫1▒〖4 . 𝑡^(− 1/2) 〗. 𝑑𝑡 = ∫1▒𝑡^(1/2) . 𝑑𝑡 − ∫1▒〖4 . 𝑡^(− 1/2) 〗. 𝑑𝑡 (As 𝑥 + 4=𝑡 ⇒𝑥=𝑡−4) = 𝑡^(1/2 + 1)/(1/2 + 1) − 4 𝑡^(− 1/2 + 1)/(− 1/2 + 1) +𝐶 = (𝑡^(3/2) )/(3/2) − 4 𝑡^(1/2)/(1/2) +𝐶 = 2/3 𝑡^(3/2) − 4 . 2𝑡^(1/2) +𝐶 Taking 2/3 . 𝑡^(1/2) as common , we get = 2/3 . 𝑡^(1/2) (𝑡−4 . 3)+𝐶 = 2/3 . 𝑡^(1/2) (𝑡−12)+𝐶 Putting the value of 𝑡=𝑥+4 = 2/3 . (𝑥+4)^(1/2) (𝑥+4−12)+𝐶 = 𝟐/𝟑 √(𝒙+𝟒) (𝒙−𝟖)+𝑪