Question 20
Evaluate: Definite Integration -1 → 1 ∫ (x + |x| + 1) / (x 2 + 2|x| + 1)




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Last updated at Oct. 1, 2019 by Teachoo
Question 20
Evaluate: Definite Integration -1 → 1 ∫ (x + |x| + 1) / (x 2 + 2|x| + 1)
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Transcript
Question 20 Evaluate: β«1_(β1)^1β(π₯+|π₯|+1)/(π₯^2+2|π₯|+1) We know that |π₯|={β(β&π₯, π₯<0@&π₯, π₯β₯0)β€ So, for β1 to 0, |x| = βx and for 0 to 1 |x| = x So, our integral becomes β«1_(β1)^1β(π₯+|π₯|+1)/(π₯^2+2|π₯|+1) β«1_(β1)^0β(π₯+|π₯|+1)/(π₯^2+2|π₯|+1) β«1_0^1β(π₯+|π₯|+1)/(π₯^2+2|π₯|+1) β«1_(β1)^0β(π₯βπ₯+1)/(π₯^2β2π₯+1) β«1_0^1β(π₯+π₯+1)/(π₯^2+2π₯+1) β«1_(β1)^0β1/(π₯^2β2π₯+1) β«1_0^1β(2π₯+1)/(π₯^2+2π₯+1) β«1_(β1)^0β1/(π₯β1)^2 β«1_0^1β(2π₯+1)/(π₯+1)^2 β«1_(β1)^0β1/(π₯β1)^2 β«1_0^1β(2π₯+1+1β1)/(π₯+1)^2 β«1_(β1)^0β1/(π₯β1)^2 β«1_0^1β(2π₯+2β1)/(π₯+1)^2 β«1_(β1)^0β1/(π₯β1)^2 β«1_0^1β(2(π₯+1)β1)/(π₯+1)^2 β«1_(β1)^0β1/(π₯β1)^2 β«1_0^1β(2(π₯+1) )/(π₯+1)^2 β«1_0^1β1/(π₯+1)^2 β«1_(β1)^0β1/(π₯β1)^2 β«1_0^1β(2 )/((π₯+1) ) β«1_0^1β1/(π₯+1)^2 β«1_(β1)^0β1/(π₯β1)^2 2 β«1_0^1β"dx " /((π₯+1) ) β«1_0^1β1/(π₯+1)^2 [(π₯β1)^(β1)/(β1)]_(β1)^0 2[lnβ‘γ|π₯+1|γ ]_0^1 [(π₯+1)^(β1)/(β1)]_0^1 [(β1)/((π₯β1))]_(β1)^0 "2" [lnβ‘γ|π₯+1|γ ]_0^1 [(β1)/((π₯+1))]_0^1 [(β1)/((0β1))β(β1)/((β1β1))] 2[lnβ‘γ|2|γβlogβ‘γ|1|γ ] [(β1)/((1+1))β(β1)/((0+1))] [1β1/2] 2[lnβ‘γ|2|γβ0] [β1/2+1] 1/2+2 lnβ‘2β1/2 2 lnβ‘2
CBSE Class 12 Sample Paper for 2019 Boards
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