Question

How do calculate that? pls help me

`int(x^2+2x+3)/(x^2+2x+1) dx`

Answer 1

`int(x^2+2x+3)/(x^2+2x+1)dx=x-2/(x+1)+c`

Explanation:

`int(x^2+2x+3)/(x^2+2x+1) dx`

= `int[(x^2+2x+1)/(x^2+2x+1)+2/(x^2+2x+1)]dx`

as `x^2+2x+1=(x+1)^2` this becomes

`int(1+2/(x+1)^2)dx`

= `intdx+2int1/(x+1)^2dx`

= `x+2int1/(x+1)^2` - now let `u=x+1` then `du=dx`

and our integral becomes

`x+2int(du)/u^2`

= `x-2/u+c`

= `x-2/(x+1)+c`

Answer 2

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

Explanation:

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

`\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` and simply drop the `1` because it's a constant:

`\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\int1/(x^2+2x+1)dx=2\int1/(x+1)^2dx+\int 1dx`

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

`2\int1/(x+1)^2dx+\int 1dx=2\int1/(x+1)^2dx+x`

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

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

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

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