Similarly for the following integrals a) I = 2x2+3x-1 dx x3+7x2 +14x+8 To factor the denominator,...

 

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Similarly for the following integrals a) I = 2x2+3x-1 dx x3+7x2 +14x+8 To factor the denominator, we more challenging x²+3x+8 x3+7x2+7x-15 polynomial division well B the results: polynomial then: a) If P(x) is a (x-a) \ P(x) = P(a) = 0 8) If P(x) = x + an-1x4-1 t... faix tao then (P(p)=0 ApeQ) → plaa. c) I= I { 2x2+3 dx dx (x²-1)(x2+1) + Because the integral is exclusively dependent we can use the dh x², Paravartya-Sutra method to expand 2913 A B + y-1 y+ 1 (y-1)(y+1) and then set y=x². Similarly for the following integrals a) I = 2x2+3x-1 dx x3+7x2 +14x+8 To factor the denominator, we more challenging x²+3x+8 x3+7x2+7x-15 polynomial division well B the results: polynomial then: a) If P(x) is a (x-a) \ P(x) = P(a) = 0 8) If P(x) = x + an-1x4-1 t... faix tao then (P(p)=0 ApeQ) → plaa. c) I= I { 2x2+3 dx dx (x²-1)(x2+1) + Because the integral is exclusively dependent we can use the dh x², Paravartya-Sutra method to expand 2913 A B + y-1 y+ 1 (y-1)(y+1) and then set y=x².

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a To integrate the expression 2x 3x 1 x 7x 14x 8 we can first perform polynomial division 2x 4 x 7x ... View the full answer