Question 20

Evaluate: Definite Integration -1 → 1  ∫ (x + |x| + 1) / (x + 2|x| + 1)

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Question 20 Evaluate: ∫1_(βˆ’1)^1β–’(π‘₯+|π‘₯|+1)/(π‘₯^2+2|π‘₯|+1) dx We know that |π‘₯|={β–ˆ(βˆ’&π‘₯, π‘₯<0@&π‘₯, π‘₯β‰₯0)─ So, for –1 to 0, |x| = –x and for 0 to 1 |x| = x So, our integral becomes ∫1_(βˆ’1)^1β–’(π‘₯+|π‘₯|+1)/(π‘₯^2+2|π‘₯|+1) ∫1_(βˆ’1)^0β–’(π‘₯+|π‘₯|+1)/(π‘₯^2+2|π‘₯|+1) ∫1_0^1β–’(π‘₯+|π‘₯|+1)/(π‘₯^2+2|π‘₯|+1) ∫1_(βˆ’1)^0β–’(π‘₯βˆ’π‘₯+1)/(π‘₯^2βˆ’2π‘₯+1) ∫1_0^1β–’(π‘₯+π‘₯+1)/(π‘₯^2+2π‘₯+1) ∫1_(βˆ’1)^0β–’1/(π‘₯^2βˆ’2π‘₯+1) ∫1_0^1β–’(2π‘₯+1)/(π‘₯^2+2π‘₯+1) ∫1_(βˆ’1)^0β–’1/(π‘₯βˆ’1)^2 ∫1_0^1β–’(2π‘₯+1)/(π‘₯+1)^2 ∫1_(βˆ’1)^0β–’1/(π‘₯βˆ’1)^2 ∫1_0^1β–’(2π‘₯+1+1βˆ’1)/(π‘₯+1)^2 ∫1_(βˆ’1)^0β–’1/(π‘₯βˆ’1)^2 ∫1_0^1β–’(2π‘₯+2βˆ’1)/(π‘₯+1)^2 ∫1_(βˆ’1)^0β–’1/(π‘₯βˆ’1)^2 ∫1_0^1β–’(2(π‘₯+1)βˆ’1)/(π‘₯+1)^2 ∫1_(βˆ’1)^0β–’1/(π‘₯βˆ’1)^2 ∫1_0^1β–’(2(π‘₯+1) )/(π‘₯+1)^2 ∫1_0^1β–’1/(π‘₯+1)^2 Let (x + 1)2 = t 2 (x+ 1) dx = dt So, when x β†’ 0 , t β†’ 12 = 1 and x β†’ 1, t β†’ 22 = 4 ∫1_(βˆ’1)^0β–’1/(π‘₯βˆ’1)^2 ∫1_1^4▒𝑑𝑑/𝑑 ∫1_0^1β–’1/(π‘₯+1)^2 [(π‘₯βˆ’1)^(βˆ’1)/(βˆ’1)]_(βˆ’1)^0 [ln⁑〖|𝑑|γ€— ]_1^4 [(π‘₯+1)^(βˆ’1)/(βˆ’1)]_0^1 [(βˆ’1)/((π‘₯βˆ’1))]_(βˆ’1)^0 [ln⁑〖|𝑑|γ€— ]_1^4 [(βˆ’1)/((π‘₯+1))]_0^1 [(βˆ’1)/((0βˆ’1))βˆ’(βˆ’1)/((βˆ’1βˆ’1))] [ln⁑〖|4|γ€—βˆ’log⁑〖|1|γ€— ] [(βˆ’1)/((1+1))βˆ’(βˆ’1)/((0+1))] [1βˆ’1/2] [ln⁑〖|4|γ€—βˆ’0] [βˆ’1/2+1] 1/2+ln⁑4βˆ’1/2 ln⁑4 ln⁑〖2^2 γ€— 2ln⁑2

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.