How do you integrate $\int \frac{x^{2} + 2 x}{x^{2} + 2 x + 1}$ $int (x^2+2x)/(x^2+2x+1)$ using substitution?

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Answer 1 · 2017-02-13T08:04:28

The answer is $= (x + 1) + \frac{1}{x + 1} + C$ $=(x+1)+1/(x+1)+C$

Let's factorise,

Therefore,

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

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

Let $u = x + 1$ $u=x+1$

$d u = d x$ $du=dx$

$x + 2 = u + 1$ $x+2=u+1$

$x = u - 1$ $x=u-1$

So,

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

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

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

$= u + \frac{1}{u}$ $=u+1/u$

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

Answer 2 · 2017-02-13T17:08:09

I have found the following results with one substitution:
enter image source here

How do you integrate int (x^2+2x)/(x^2+2x+1)∫x2+2xx2+2x+1int (x^2+2x)/(x^2+2x+1) using substitution?