∫x^2+x+1/(x+1)^2(x+2)dx The integrand x^2+x+1/(x+1)^2(x+2)is a proper rational function. ∴ x^2+x+1/(x+1)^2(x+2)=A/x+1+B/(x+1)^2+C/(x+2) ....(i) ⇒ x^2+x+ ...

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Algebra of Limits

∫x2+x+1/x+12x...

Question

$\int \frac{x^{2} + x + 1}{(x + 1)^{2} (x + 2)} d x .$

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Solution

$\int \frac{x^{2} + x + 1}{(x + 1)^{2} (x + 2)} d x T h e i n t e g r a n d \frac{x^{2} + x + 1}{(x + 1)^{2} (x + 2)} i s a p r o p e r r a t i o n a l f u n c t i o n . ∴ \frac{x^{2} + x + 1}{(x + 1)^{2} (x + 2)} = \frac{A}{x + 1} + \frac{B}{(x + 1)^{2}} + \frac{C}{(x + 2)} . . . . (i) \Rightarrow x^{2} + x + 1 = A (x + 1) (x + 2) + B (x + 2) + C (x + 1)^{2} \Rightarrow x^{2} + x + 1 = A (x^{2} + 3 x + 2) + B (x + 2) + c (x^{2} + 2 x + 1) \Rightarrow x^{2} + x + 1 = (A + C) x^{2} + (3 A + B + 2 C) x + (2 A + 2 B + C)$

On comparing the coefficients of like powers on both sides, we get

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

On solving these equations, we get

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

$F r o m E q . (i), w e g e t \frac{x^{2} + x + 1}{(x + 1)^{2} (x + 2)} = \frac{- 2}{x + 1} + \frac{1}{(x + 1)^{2}} + \frac{3}{(x + 2)} ∴ \frac{x^{2} + x + 1}{(x + 1)^{2} (x + 2)} d x = - 2 \int \frac{d x}{x + 1} = \int \frac{d x}{(x + 1)^{2}} + 3 \int \frac{d x}{(x + 2)} = - 2 l o g | x + 1 | - \frac{1}{(x + 1)} + 3 l o g | x + 2 | + C$

Suggest Corrections

∫x2+x+1(x+1)2(x+2)dx.

$\int \frac{x^{2} + x + 1}{(x + 1)^{2} (x + 2)} d x .$