∫ 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
Last updated at Dec. 14, 2024 by Teachoo
This question is similar to Question 20 - CBSE Class 12 - Sample Paper for 2019 Board
Question 7 โซ1_(โ1)^1โใ (๐ฅ^3 + |๐ฅ| + 1)/(๐ฅ^2 + 2|๐ฅ| + 1)ใ ๐๐ฅ is equal to (A) logโก2 (B) 2 logโก2 (C) 1/2 logโก2 (D) 4 logโก2 We know that |๐ฅ|={โ(โ&๐ฅ, ๐ฅ<0@&๐ฅ, ๐ฅโฅ0)โค So, for โ1 to 0, |x| = โx and for 0 to 1 |x| = x So, our integral becomes โซ1_(โ1)^1โ(๐ฅ+|๐ฅ|+1)/(๐ฅ^2+2|๐ฅ|+1) โซ1_(โ1)^0โ(๐ฅ+|๐ฅ|+1)/(๐ฅ^2+2|๐ฅ|+1) โซ1_0^1โ(๐ฅ+|๐ฅ|+1)/(๐ฅ^2+2|๐ฅ|+1) โซ1_(โ๐)^๐โ(๐โ๐+๐)/(๐^๐โ๐๐+๐) โซ1_๐^๐โ(๐+๐+๐)/(๐^๐+๐๐+๐) โซ1_(โ1)^0โ1/(๐ฅ^2โ2๐ฅ+1) โซ1_0^1โ(2๐ฅ+1)/(๐ฅ^2+2๐ฅ+1) โซ1_(โ๐)^๐โ๐/(๐โ๐)^๐ โซ1_๐^๐โ(๐๐+๐)/(๐+๐)^๐ โซ1_(โ1)^0โ1/(๐ฅโ1)^2 โซ1_0^1โ(2๐ฅ+1+1โ1)/(๐ฅ+1)^2 โซ1_0^1โ(2๐ฅ+1+1โ1)/(๐ฅ+1)^2 โซ1_(โ1)^0โ1/(๐ฅโ1)^2 โซ1_0^1โ(2๐ฅ+2โ1)/(๐ฅ+1)^2 โซ1_(โ๐)^๐โ๐/(๐โ๐)^๐ โซ1_๐^๐โ(๐(๐+๐)โ๐)/(๐+๐)^๐ โซ1_(โ1)^0โ1/(๐ฅโ1)^2 โซ1_0^1โ(2(๐ฅ+1) )/(๐ฅ+1)^2 โซ1_0^1โ1/(๐ฅ+1)^2 โซ1_(โ1)^0โ1/(๐ฅโ1)^2 โซ1_๐^๐โ(๐ )/((๐+๐) ) โซ1_๐^๐โ๐/(๐+๐)^๐ โซ1_(โ1)^0โ1/(๐ฅโ1)^2 2 โซ1_0^1โ"dx " /((๐ฅ+1) ) โซ1_0^1โ1/(๐ฅ+1)^2 [(๐ฅโ1)^(โ1)/(โ1)]_(โ1)^0 2[logโกใ|๐ฅ+1|ใ ]_0^1 [(๐ฅ+1)^(โ1)/(โ1)]_0^1 1/2+2 logโก2โ1/2 2 ๐ฅ๐จ๐ โก๐ So, the correct answer is (b)
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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo