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7. Integrals
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Q22 of 271 Page 352

Integrate the function:

Given:

Let


Using partial differentiation:


let




⇒ x2 + x + 1 = Ax2 + 3Ax + 2A + Bx +2B + Cx2 + 2Cx + C


⇒ x2 + x + 1 = (2A+2B+C) + (3A+B+2C)x + (A+C)x2


Equating the coefficients of x, x2 and constant value. We get:


(a) A + C = 1


(b) 3A + B + 2C = 1


( c) 2A+2B+C =1


After solving we get:


A=-2, B=1 and C=3








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Questions · 271
7. Integrals
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