Question 20 - Evaluate definite integral -1 -> 1 (x + |x| + 1) / x^2 +

Question 20 Evaluate: ∫1_(−1)^1▒(𝑥+|𝑥|+1)/(𝑥^2+2|𝑥|+1) dx 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 Let (x + 1)2 = t 2 (x+ 1) dx = dt So, when x → 0 , t → 12 = 1 and x → 1, t → 22 = 4 ∫1_(−1)^0▒1/(𝑥−1)^2 ∫1_1^4▒𝑑𝑡/𝑡 ∫1_0^1▒1/(𝑥+1)^2 [(𝑥−1)^(−1)/(−1)]_(−1)^0 [ln⁡〖|𝑡|〗 ]_1^4 [(𝑥+1)^(−1)/(−1)]_0^1 [(−1)/((𝑥−1))]_(−1)^0 [ln⁡〖|𝑡|〗 ]_1^4 [(−1)/((𝑥+1))]_0^1 [(−1)/((0−1))−(−1)/((−1−1))] [ln⁡〖|4|〗−log⁡〖|1|〗 ] [(−1)/((1+1))−(−1)/((0+1))] [1−1/2] [ln⁡〖|4|〗−0] [−1/2+1] 1/2+ln⁡4−1/2 ln⁡4 ln⁡〖2^2 〗 2ln⁡2

Question 20 - CBSE Class 12 Sample Paper for 2019 Boards