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‘₯<[email protected]&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 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.