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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Question 7 โˆซ1_(โˆ’1)^1โ–’ใ€– (๐‘ฅ^3 + |๐‘ฅ| + 1)/(๐‘ฅ^2 + 2|๐‘ฅ| + 1)ใ€— ๐‘‘๐‘ฅ is equal to (A) logโก2 (B) 2 logโก2 (C) 1/2 logโก2 (D) 4 logโก2 We know that |๐‘ฅ|={โ–ˆ(โˆ’&๐‘ฅ, ๐‘ฅ<0@&๐‘ฅ, ๐‘ฅโ‰ฅ0)โ”ค So, for โ€“1 to 0, |x| = โ€“x and for 0 to 1 |x| = x So, our integral becomes โˆซ1_(โˆ’1)^1โ–’(๐‘ฅ+|๐‘ฅ|+1)/(๐‘ฅ^2+2|๐‘ฅ|+1) โˆซ1_(โˆ’1)^0โ–’(๐‘ฅ+|๐‘ฅ|+1)/(๐‘ฅ^2+2|๐‘ฅ|+1) โˆซ1_0^1โ–’(๐‘ฅ+|๐‘ฅ|+1)/(๐‘ฅ^2+2|๐‘ฅ|+1) โˆซ1_(โˆ’๐Ÿ)^๐ŸŽโ–’(๐’™โˆ’๐’™+๐Ÿ)/(๐’™^๐Ÿโˆ’๐Ÿ๐’™+๐Ÿ) โˆซ1_๐ŸŽ^๐Ÿโ–’(๐’™+๐’™+๐Ÿ)/(๐’™^๐Ÿ+๐Ÿ๐’™+๐Ÿ) โˆซ1_(โˆ’1)^0โ–’1/(๐‘ฅ^2โˆ’2๐‘ฅ+1) โˆซ1_0^1โ–’(2๐‘ฅ+1)/(๐‘ฅ^2+2๐‘ฅ+1) โˆซ1_(โˆ’๐Ÿ)^๐ŸŽโ–’๐Ÿ/(๐’™โˆ’๐Ÿ)^๐Ÿ โˆซ1_๐ŸŽ^๐Ÿโ–’(๐Ÿ๐’™+๐Ÿ)/(๐’™+๐Ÿ)^๐Ÿ โˆซ1_(โˆ’1)^0โ–’1/(๐‘ฅโˆ’1)^2 โˆซ1_0^1โ–’(2๐‘ฅ+1+1โˆ’1)/(๐‘ฅ+1)^2 โˆซ1_0^1โ–’(2๐‘ฅ+1+1โˆ’1)/(๐‘ฅ+1)^2 โˆซ1_(โˆ’1)^0โ–’1/(๐‘ฅโˆ’1)^2 โˆซ1_0^1โ–’(2๐‘ฅ+2โˆ’1)/(๐‘ฅ+1)^2 โˆซ1_(โˆ’๐Ÿ)^๐ŸŽโ–’๐Ÿ/(๐’™โˆ’๐Ÿ)^๐Ÿ โˆซ1_๐ŸŽ^๐Ÿโ–’(๐Ÿ(๐’™+๐Ÿ)โˆ’๐Ÿ)/(๐’™+๐Ÿ)^๐Ÿ โˆซ1_(โˆ’1)^0โ–’1/(๐‘ฅโˆ’1)^2 โˆซ1_0^1โ–’(2(๐‘ฅ+1) )/(๐‘ฅ+1)^2 โˆซ1_0^1โ–’1/(๐‘ฅ+1)^2 โˆซ1_(โˆ’1)^0โ–’1/(๐‘ฅโˆ’1)^2 โˆซ1_๐ŸŽ^๐Ÿโ–’(๐Ÿ )/((๐’™+๐Ÿ) ) โˆซ1_๐ŸŽ^๐Ÿโ–’๐Ÿ/(๐’™+๐Ÿ)^๐Ÿ โˆซ1_(โˆ’1)^0โ–’1/(๐‘ฅโˆ’1)^2 2 โˆซ1_0^1โ–’"dx " /((๐‘ฅ+1) ) โˆซ1_0^1โ–’1/(๐‘ฅ+1)^2 [(๐‘ฅโˆ’1)^(โˆ’1)/(โˆ’1)]_(โˆ’1)^0 2[logโกใ€–|๐‘ฅ+1|ใ€— ]_0^1 [(๐‘ฅ+1)^(โˆ’1)/(โˆ’1)]_0^1 1/2+2 logโก2โˆ’1/2 2 ๐ฅ๐จ๐ โก๐Ÿ 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.