Integrate x-1/(x+1)²?

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Answer 1 · 2018-05-13T11:44:53

#int (x-1/(x+1)^2)dx =(x^3+x+2)/(2(x+1))+C#

Using the linearity of the integral:

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

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

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

#int (x-1/(x+1)^2)dx =(x^3+x+2)/(2(x+1))+C#

Answer 2 · 2018-05-13T14:19:38

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

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#.