How do calculate that? pls help me

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

2 Answers
Apr 14, 2018

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

Apr 14, 2018

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