Question 20

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

Find Integration  (x + |x| + 1) / (x2 + 2|x| + 1) from -1 to 1

Question 20 - CBSE Class 12 Sample Paper for 2019 Boards - Part 2
Question 20 - CBSE Class 12 Sample Paper for 2019 Boards - Part 3 Question 20 - CBSE Class 12 Sample Paper for 2019 Boards - Part 4 Question 20 - CBSE Class 12 Sample Paper for 2019 Boards - Part 5

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Transcript

Question 20 Evaluate: ∫1_(−1)^1▒(𝑥+|𝑥|+1)/(𝑥^2+2|𝑥|+1) We know that |𝑥|={█(−&𝑥, 𝑥<0@&𝑥, 𝑥≥0)┤ So, for –1 to 0, |x| = –x and for 0 to 1 |x| = x So, our integral becomes ∫1_(−1)^1▒(𝑥+|𝑥|+1)/(𝑥^2+2|𝑥|+1) ∫1_(−1)^0▒(𝑥+|𝑥|+1)/(𝑥^2+2|𝑥|+1) ∫1_0^1▒(𝑥+|𝑥|+1)/(𝑥^2+2|𝑥|+1) ∫1_(−1)^0▒(𝑥−𝑥+1)/(𝑥^2−2𝑥+1) ∫1_0^1▒(𝑥+𝑥+1)/(𝑥^2+2𝑥+1) ∫1_(−1)^0▒1/(𝑥^2−2𝑥+1) ∫1_0^1▒(2𝑥+1)/(𝑥^2+2𝑥+1) ∫1_(−1)^0▒1/(𝑥−1)^2 ∫1_0^1▒(2𝑥+1)/(𝑥+1)^2 ∫1_(−1)^0▒1/(𝑥−1)^2 ∫1_0^1▒(2𝑥+1+1−1)/(𝑥+1)^2 ∫1_(−1)^0▒1/(𝑥−1)^2 ∫1_0^1▒(2𝑥+2−1)/(𝑥+1)^2 ∫1_(−1)^0▒1/(𝑥−1)^2 ∫1_0^1▒(2(𝑥+1)−1)/(𝑥+1)^2 ∫1_(−1)^0▒1/(𝑥−1)^2 ∫1_0^1▒(2(𝑥+1) )/(𝑥+1)^2 ∫1_0^1▒1/(𝑥+1)^2 ∫1_(−1)^0▒1/(𝑥−1)^2 ∫1_0^1▒(2 )/((𝑥+1) ) ∫1_0^1▒1/(𝑥+1)^2 ∫1_(−1)^0▒1/(𝑥−1)^2 2 ∫1_0^1▒"dx " /((𝑥+1) ) ∫1_0^1▒1/(𝑥+1)^2 [(𝑥−1)^(−1)/(−1)]_(−1)^0 2[ln⁡〖|𝑥+1|〗 ]_0^1 [(𝑥+1)^(−1)/(−1)]_0^1 [(−1)/((𝑥−1))]_(−1)^0 "2" [ln⁡〖|𝑥+1|〗 ]_0^1 [(−1)/((𝑥+1))]_0^1 [(−1)/((0−1))−(−1)/((−1−1))] 2[ln⁡〖|2|〗−log⁡〖|1|〗 ] [(−1)/((1+1))−(−1)/((0+1))] [1−1/2] 2[ln⁡〖|2|〗−0] [−1/2+1] 1/2+2 ln⁡2−1/2 2 ln⁡2

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.