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

(A) logโก2ย 

(B) 2 logโก2ย 

(C) 1/2 ย logโก2ย 

(D) 4 logโก2

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

[Integrals Class 12 MCQ] - Find Intgeral from -1 to 1  โˆซ (๐‘ฅ^3 + |๐‘ฅ| - NCERT Exemplar MCQ

part 2 - Question 7 - NCERT Exemplar MCQ - Serial order wise - Chapter 7 Class 12 Integrals
part 3 - Question 7 - NCERT Exemplar MCQ - Serial order wise - Chapter 7 Class 12 Integrals part 4 - Question 7 - NCERT Exemplar MCQ - Serial order wise - Chapter 7 Class 12 Integrals part 5 - Question 7 - NCERT Exemplar MCQ - Serial order wise - Chapter 7 Class 12 Integrals

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Transcript

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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