Intregration

Asked by thebluebloo | 12th 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  | 12th Feb, 2010, 07:11: PM

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