Check sibling questions

∫ 1^(-1) (x^3 + |x| + 1) / (x^2 + 2|x| + 1) dx is equal to

(A) log⁑2 

(B) 2 log⁑2 

(C) 1/2  log⁑2 

(D) 4 log⁑2

This question is similar to Question 20 - CBSE Class 12 - Sample Paper for 2019 Board

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Transcript

Question 7 ∫1_(βˆ’1)^1β–’γ€– (π‘₯^3 + |π‘₯| + 1)/(π‘₯^2 + 2|π‘₯| + 1)γ€— 𝑑π‘₯ is equal to (A) log⁑2 (B) 2 log⁑2 (C) 1/2 log⁑2 (D) 4 log⁑2 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’πŸ)^πŸŽβ–’(π’™βˆ’π’™+𝟏)/(𝒙^πŸβˆ’πŸπ’™+𝟏) ∫1_𝟎^πŸβ–’(𝒙+𝒙+𝟏)/(𝒙^𝟐+πŸπ’™+𝟏) ∫1_(βˆ’1)^0β–’1/(π‘₯^2βˆ’2π‘₯+1) ∫1_0^1β–’(2π‘₯+1)/(π‘₯^2+2π‘₯+1) ∫1_(βˆ’πŸ)^πŸŽβ–’πŸ/(π’™βˆ’πŸ)^𝟐 ∫1_𝟎^πŸβ–’(πŸπ’™+𝟏)/(𝒙+𝟏)^𝟐 ∫1_(βˆ’1)^0β–’1/(π‘₯βˆ’1)^2 ∫1_0^1β–’(2π‘₯+1+1βˆ’1)/(π‘₯+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’πŸ)^πŸŽβ–’πŸ/(π’™βˆ’πŸ)^𝟐 ∫1_𝟎^πŸβ–’(𝟐(𝒙+𝟏)βˆ’πŸ)/(𝒙+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 ∫1_(βˆ’1)^0β–’1/(π‘₯βˆ’1)^2 ∫1_𝟎^πŸβ–’(𝟐 )/((𝒙+𝟏) ) ∫1_𝟎^πŸβ–’πŸ/(𝒙+𝟏)^𝟐 ∫1_(βˆ’1)^0β–’1/(π‘₯βˆ’1)^2 2 ∫1_0^1β–’"dx " /((π‘₯+1) ) ∫1_0^1β–’1/(π‘₯+1)^2 [(π‘₯βˆ’1)^(βˆ’1)/(βˆ’1)]_(βˆ’1)^0 2[log⁑〖|π‘₯+1|γ€— ]_0^1 [(π‘₯+1)^(βˆ’1)/(βˆ’1)]_0^1 1/2+2 log⁑2βˆ’1/2 2 π₯𝐨𝐠⁑𝟐 So, the correct answer is (b)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.