CBSE Class 12-science Answered
Intregration
Asked by thebluebloo | 12 Feb, 2010, 09:38: AM
Expert Answer
Using partial fractions,
x/(x2+1)(x+1) = (Ax+B)/(x2+1) + C/(x+1)
x = (Ax+B)(x+1) + C(x2+1)
x = (A+C)x2 + (A+B)x +(B+C)
Comparing powers of x on both sides, we find,
A = -C; A + B = 1; B = -C
A = B = 1/2 and C = -1/2
Hence,
x dx / (x2+1) (x+1) = (1/2) [(x+1)/(x2+1) - 1/(x+1)] dx
= (1/2)[(x/(x2+1) + 1/(x2+1) - 1/(x+1)] dx
= (1/2)[(1/2)log(x2+1) + tan-1(x) - log(x+1)] + C
= log{[((x2+1))/(x+1)]} + (1/2)tan-1x + C
Regards,
Team,
TopperLearning.
Answered by | 12 Feb, 2010, 07:11: PM
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