Question #1377f

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Question #1377f

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Answer 1 · 2017-04-16T23:49:58

$2 x - \frac{1}{x - 1} + C$ $2x-1/(x-1)+C$

Explanation:

$\int \frac{2 x^{2} - 4 x + 3}{{(x - 1)}^{2}} d x$ $int(2x^2-4x+3)/(x-1)^2dx$

Let $u = x - 1$ $u=x-1$ . This implies that $x = u + 1$ $x=u+1$ and $d u = d x$ $du=dx$ . Then:

$= \int \frac{2 {(u + 1)}^{2} - 4 (u + 1) + 3}{u^{2}} d u$ $=int(2(u+1)^2-4(u+1)+3)/u^2du$

Expanding and combining like terms:

$= \int \frac{2 u^{2} + 1}{u^{2}} d u$ $=int(2u^2+1)/u^2du$

$= \int (2 + \frac{1}{u^{2}}) d u$ $=int(2+1/u^2)du$

$= 2 u - \frac{1}{u} + C$ $=2u-1/u+C$

$= 2 (x - 1) - \frac{1}{x - 1} + C$ $=2(x-1)-1/(x-1)+C$

Expanding $2 (x - 1)$ $2(x-1)$ , we see that the $- 2$ $-2$ combines with the constant of integration:

$= 2 x - \frac{1}{x - 1} + C$ $=2x-1/(x-1)+C$

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Question #1377f

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