Example 16 - Chapter 7 Class 12 Integrals
Last updated at December 16, 2024 by Teachoo
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Chapter 7 Class 12 Integrals
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Example 16 Find ā«1ā(š„^2+ š„ +1 šš„ )/((š„ + 2) (š„^2+1) ) We can write equation as (š„^2+ š„ + 1)/((š„ + 1) (š„ + 2) )=š“/(š„ + 2) + (šµš„ + š¶)/(š„^2+ 1) Cancelling denominator ć š„ć^2+ š„+1=š“(š„^2+1)+(šµš„+š¶) (š„+2) Putting x = āš (ā2)^2+(ā2)+1=š“((ā2)^2+1)+0 4ā2+1= 5A 3/5 = A Putting x = š š„^2+ š„+1=š“(š„^2+1)+(šµš„+š¶) (š„+2) 0+0+0= A(0 + 1) + (0 + C) (0 + 2) 1 = A + 2C 1 = 3/5 + 2C 1 ā 3/5 = 2C 2/5 = 2C C = 1/5 Putting x = 1 š„^2+ š„+1=š“(š„^2+1)+(šµš„+š¶) (š„+2) 1+1+1= 2A + (B + C)(3) 3 = 2A + 3 (B + C) 3 = 2(3/5) + 3 (B+1/5) 3 ā 6/5 = 3 (B+1/5) 9/5 = 3 (B+1/5) 3/5 ā 1/5 = B B = 2/5 Thus, (š„^2+ š„ + 1)/((š„ + 1) (š„ + 2) )=š“/(š„ + 2) + (šµš„ + š¶)/(š„^2+ 1) (š„^2+ š„ + 1)/((š„ + 1)(š„^2+ 1)) = 3/(5 (š„ + 2)) + (1 (2š„ + 1))/(5 (š„^2 + 1)) Hence, our equation becomes ā«1ā(š„^2+ š„ + 1)/((š„ + 2) (š„^2 + 1)) šš„= ā«1ā3/(5(š„^2 + 1)) šš„+ā«1ā1/5 ((2š„ + 1))/(š„^2 + 1) šš„ = ā«1ā3/(5(š„^2 + 1)) šš„+ 1/5 ā«1āć2š„/(š„^2 + 1) šš„+ć 1/5 ā«1ā1/(š„^2 + 1) šš„ šš 1/5 ā«1ā2š„/(š„^2+ 1) šš„ Let š”=š„^2+ 1 šš”/šš„=2š„ šš”=2š„ šš„ Substituting, =1/5 ā«1āšš”/š” = 1/5 log |š”| + C_2 = 1/5 log |š„^2+1| + C_2 šš 1/5 ā«1ā1/(š„^2+ 1) šš„ = 1/5 ćš”ššć^(ā1) (š„)+C_3 Hence ā«1ā(š„^2+ š„ + 1)/((š„ + 2) (š„^2+ 1)) šš„ =š/š ššš|š+š|+š/š ššš|š^š+š|+š/š ćšššć^(āš) (š)+ C where C = C_1+ C_2+C_3
