Mathematics Dictionary Dr. K. G. Shih

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Integral : Partial fractions

Subjects

Symbol Defintion Example : Sqr(x) is square root of x Q01 | - Partial fraction Q02 | - ∫(1/(1 - x^2))dx Q03 | - Q04 | - Q05 | - Q06 | - Q07 | - Q08 | - Q09 | - Q10 | -

Answers

Q01. Partial frction Partial fraction Subects | Partial fractin Go to Begin Q02. ∫(1/(1 - x^2))dx Express 1/(1 - x^2) = A/(1 - x) + B/(1 + x) Both sides times (1 - x^2) We have 1 = A*(1 + x) + B*(1 - x) or 1 = (A - B)*x + (A + B) Hence A - B = 0 and A + B = 1 Hence A = B = 1/2 Find integral ∫(1/(1 - x^2))dx = ∫(1/2)*(1/(1 - x) + 1/(1 + x))dx = (1/2)*(ln(1 + x) - ln(1 - x)) + C = (1/2)*ln((1 + x)/(1 - x)) + C Verify d/dx(ln(1 + x)/(1 - x)) = ((1 - x)/(1 + x))*d/dx((1 + x)/(1 - x)) = ((1 - x)/(1 + x))*((1 - x) + (1 + x))/(1 - x^2)) = ((1 - x)/(1 + x))*(2/(1 - x^2)) = 1/((1 - x)*(1 + x)) = 1/(1 - x^2) Prove that ∫(1/(1 - x^2))dx = arctanh(x) derivative : d/dx(arctnah(x) = 1/(1 - x^2) : See 06 04 Integral : See 14 09 Go to Begin Q03. Go to Begin Q04. 1. Join mid points of two sides which is parallel to other side Go to Begin Q05. Go to Begin Q06. New Go to Begin Q07. Answer Go to Begin Q08. Answer Go to Begin Q09. Answer Go to Begin Q10. Answer Go to Begin

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