Question

What is the antiderivative of `ln(x^2 + 2x + 2)`?

Answer 1

`xln(x^2+2x+2) - 2x + ln(x^2+2x+2)+2tan^-1x+C`

Explanation:

The antiderivative of `ln x` can be easily found out by integration by parts to be

`int ln x dx = ln x - int (d/dx(ln x) times int 1 *dx)dx`
`qquad = xln x-x+C`

The antiderivative of the given function is

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

`qquad = ln (x^2+2x+2)int 1 * dx`
`qquad -int [d/dx ln (x^2+2x+2) int 1.dx]dx`
`qquad = xln(x^2+2x+2) - int {x(2x+2)dx}/(x^2+2x+2)`

Now

`{2x(x+1)}/(x^2+2x+2) =2 {x^2+2x+2-x-2}/(x^2+2x+2)`
`qquad =2-(2x+4)/(x^2+2x+2)`
`qquad = 2-(2x+2)/(x^2+2x+2)-(2)/(x^2+2x+2)`

and so

`int {x(2x+2)dx}/(x^2+2x+2)`
`qquad = int [2-(2x+2)/(x^2+2x+2)-(2)/((x+1)^2+1)]dx`
`qquad = 2x- ln(x^2+2x+2)-2tan^-1x+C`

Thus the required antiderivative is

`xln(x^2+2x+2) - 2x + ln(x^2+2x+2)+2tan^-1x+C`