Integrate x-1/(x+1)²?

2 Answers
May 13, 2018

int (x-1/(x+1)^2)dx =(x^3+x+2)/(2(x+1))+C(x1(x+1)2)dx=x3+x+22(x+1)+C

Explanation:

Using the linearity of the integral:

int (x-1/(x+1)^2)dx = int xdx -int dx/(x+1)^2(x1(x+1)2)dx=xdxdx(x+1)2

int (x-1/(x+1)^2)dx =x^2/2 -int (d(x+1))/(x+1)^2(x1(x+1)2)dx=x22d(x+1)(x+1)2

int (x-1/(x+1)^2)dx =x^2/2 +1/(x+1)+C(x1(x+1)2)dx=x22+1x+1+C

int (x-1/(x+1)^2)dx =(x^3+x+2)/(2(x+1))+C(x1(x+1)2)dx=x3+x+22(x+1)+C

May 13, 2018

int \frac{x - 1}{( x + 1 )^2}\ d x = ln | x + 1 | + \frac{2}{x + 1} + C

Explanation:

To start with, we can rearrange the enumerator as follows:

int \frac{x - 1}{( x + 1 )^2}\ d x =

= int \frac{( x + 1 ) - 2}{( x + 1 )^2}\ d x.

Now, we may interpret x + 1 as an inner function, and thus this is the same as

[ int \frac{t - 2}{t^2}\ d t ]_{t = x + 1} =

= [ int \frac{1}{t}\ d t - 2 int \frac{1}{t^2}\ d t ]_{t = x + 1} =

= [ ln | t | + \frac{2}{t} + C ]_{t = x + 1} =

= ln | x + 1 | + \frac{2}{x + 1} + C.