Example 13 - 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 13 Find ā«1ā(3š„ ā2)/((š„ + 1)^2 (š„ + 3) ) šš„ We can write Integral as (3š„ ā 2)/((š„ + 1)^2 (š„ + 3) )=š“/(š„ + 1) + šµ/(š„ + 1)^2 + š¶/((š„ + 3) ) (3š„ ā 2)/((š„ + 1)^2 (š„ + 3) )=(š“(š„ + 1)(š„ + 3) + šµ(š„ + 3) + š¶(š„ + 1)^2)/((š„ + 1)^2 (š„ + 3) ) Cancelling denominator 3š„ ā2=š“(š„+1)(š„+3)+šµ(š„+3)+š¶(š„+1)^2 Putting x = ā1 3(ā1) ā2=š“(ā1+1)(ā1+3)+šµ(ā1+3)+š¶(ā1+1)^2 ā3ā2=š“Ć0+šµĆ2+š¶Ć(0)^2 ā5=šµĆ2 šµ=(ā 5)/2 Putting x = ā 3 3š„ ā2=š“(š„+1)(š„+3)+šµ(š„+3)+š¶(š„+1)^2 3(ā3)ā2=š“(ā3+1)(ā3+3)+šµ(ā3+3)+š¶(ā3+1)^2 ā9ā2=š“Ć0+šµĆ0+š¶Ć(ā2)^2 ā11=0+0+š¶(4) ā11=4š¶ (ā11)/4 =š¶ š¶ =(ā11)/4 Putting x = 0 3š„ ā2=š“(š„+1)(š„+3)+šµ(š„+3)+š¶(š„+1)^2 3(0) ā 2 = A(1) (3) + B(3) + C ć"(1)" ć^2 ā2 = 3A + 3B + C Putting value of B & C ā2 = 3A + 3((ā5)/2) + ((ā11)/4) ā2 = 3A ā 15/2ā11/4 ā2 = 3A + (ā30 ā 11)/4 ā8 = 12A ā 41 41 ā 8 = 12A 33/12 = A 11/4 = A Hence, we can write (3š„ ā 2)/((š„ + 1)^2 (š„ + 3) )=š“/(š„ + 1) + šµ/(š„ + 1)^2 + š¶/((š„ + 3) ) (3š„ ā 2)/((š„ + 1)^2 (š„ + 3) )=11/4(š„ + 1) ā 5/ć2(š„ + 1)ć^2 ā 11/4(š„ + 3) Therefore ā«1ā(3š„ ā 2)/((š„ + 1)^2 (š„ + 3) ) šš„ =ā«1ā11/4(š„ + 1) šš„āā«1ā5/ć2(š„ + 1)ć^2 šš„āā«1ā11/4(š„ + 3) šš„ =11/4 logā”|š„+1|ā5/2Ć((ā1))/((š„ + 1) ) ā 11/4 logā”|š„+3|+š¶ =11/4 (logā”|š„+1|ālogā”|š„+3| )+5/(2 (š„ + 1) )+š¶ =šš/š šššā”|(š + š)/(š + š)| + š/(š (š + š) )+šŖ
