How do calculate that? pls help me

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How do calculate that? pls help me

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

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Answer 1 · 2018-04-14T12:22:20

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

Explanation:

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

= $\int [\frac{x^{2} + 2 x + 1}{x^{2} + 2 x + 1} + \frac{2}{x^{2} + 2 x + 1}] d x$ $int[(x^2+2x+1)/(x^2+2x+1)+2/(x^2+2x+1)]dx$

as $x^{2} + 2 x + 1 = {(x + 1)}^{2}$ $x^2+2x+1=(x+1)^2$ this becomes

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

= $\int d x + 2 \int \frac{1}{{(x + 1)}^{2}} d x$ $intdx+2int1/(x+1)^2dx$

= $x + 2 \int \frac{1}{{(x + 1)}^{2}}$ $x+2int1/(x+1)^2$ - now let $u = x + 1$ $u=x+1$ then $d u = d x$ $du=dx$

and our integral becomes

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

= $x - \frac{2}{u} + c$ $x-2/u+c$

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

Answer 2 · 2018-04-14T12:46:04

The answer should be $x - \frac{2}{x + 1} + C$ $x-\frac{2}{x+1}+C$ .

Explanation:

First off, split the $3$ $3$ into $2 + 1$ $2+1$ :

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

Then, use the linearity of this function to get rid of the $2$ $2$ and simply drop the $1$ $1$ because it's a constant:

$\int (\frac{2}{x^{2} + 2 x + 1} + 1) d x = 2 \int \frac{1}{x^{2} + 2 x + 1} d x + \int 1 d x$ $\int(2/(x^2+2x+1)+1)dx=2\int1/(x^2+2x+1)dx+\int 1dx$

Then, try to factor out the denominator:

$2 \int \frac{1}{x^{2} + 2 x + 1} d x = 2 \int \frac{1}{{(x + 1)}^{2}} d x + \int 1 d x$ $2\int1/(x^2+2x+1)dx=2\int1/(x+1)^2dx+\int 1dx$

Get rid of $\int 1 d x$ $\int 1dx$ , using the rule $\int K d x = K x + C$ $\int Kdx=Kx+C$ , where $K$ $K$ is any constant:

$2 \int \frac{1}{{(x + 1)}^{2}} d x + \int 1 d x = 2 \int \frac{1}{{(x + 1)}^{2}} d x + x$ $2\int1/(x+1)^2dx+\int 1dx=2\int1/(x+1)^2dx+x$

Then, let $k = (x + 1)$ $k=(x+1)$ and apply the power rule:

$2 \int \frac{1}{{(x + 1)}^{2}} d x + x = 2 \int \frac{1}{k^{2}} d k + x = 2 \int k^{- 2} d k + x$ $2\int1/(x+1)^2dx+x=2\int1/k^2dk+x=2\int k^{-2}dk+x$

$2 \int k^{- 2} d k + x = 2 \frac{k^{- 2 + 1}}{- 2 + 1} + x = 2 \frac{k^{- 1}}{- 1} + x$ $2\int k^{-2}dk+x=2\frac{k^{-2+1}}{-2+1}+x=2\frac{k^{-1}}{-1}+x$

$2 \frac{k^{- 1}}{- 1} + x = \frac{- 2}{k} + x = \frac{- 2}{x + 1} + x$ $2\frac{k^{-1}}{-1}+x=\frac{-2}{k}+x=\frac{-2}{x+1}+x$

And we're done! Don't forget to add a constant, though :) Hence, we can conclude that:

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

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How do calculate that? pls help me

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

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Explanation:

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