2x-2-3x-3-x-1-x-2-2x-5-dx-

Question Number 125440 by bramlexs22 last updated on 11/Dec/20

\int \frac{2 x^{2} - 3 x - 3}{(x - 1) (x^{2} - 2 x + 5)} d x

Answered by Ar Brandon last updated on 11/Dec/20

f (x) = \frac{{2 x}^{2} - 3 x - 3}{(x - 1) (x^{2} - 2 x + 5)} = \frac{a}{x - 1} + \frac{bx + c}{x^{2} - 2 x + 5}

= \frac{a (x^{2} - 2 x + 5) + (bx + c) (x - 1)}{(x - 1) (x^{2} - 2 x + 5)}

set x - 1 = 0 \Rightarrow - 4 = 4 a \Rightarrow a = - 1

Comparing coefficients of x^{2} we get;

a + b = 2 \Rightarrow b = 3

Comparing constant terms we get;

5 a - c = - 3 \Rightarrow c = - 2

\Rightarrow f (x) = \frac{- 1}{x - 1} + \frac{3 x - 2}{x^{2} - 2 x + 5}

3 x - 2 = λ {\frac{d}{dx} (x^{2} - 2 x + 5)} + μ = λ (2 x - 2) + μ

Comparing coefs 2 λ = 3 \Rightarrow λ = \frac{3}{2}, μ - 2 λ = - 2 \Rightarrow μ = 1

3 x - 2 = \frac{3}{2} (2 x - 2) + 1

\Rightarrow \int f (x) dx = \int {\frac{- 1}{x - 1} + \frac{3}{2} \cdot \frac{2 x - 2}{x^{2} - 2 x + 5} + \frac{1}{x^{2} - 2 x + 5}} dx

Commented by Dwaipayan Shikari last updated on 11/Dec/20

\int - \frac{1}{x - 1} + \frac{3}{2} \int \frac{2 x - 2}{x^{2} - 2 x + 5} + \int \frac{1}{x^{2} - 2 x + 5} d x

= \frac{3}{2} l o g (x^{2} - 2 x + 5) - l o g (x - 1) + \frac{1}{2} {t a n}^{- 1} \frac{x - 1}{2} + C

Commented by Ar Brandon last updated on 11/Dec/20

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Answered by john_santu last updated on 11/Dec/20