How do you evaluate the integral of x2+2xdx?

1 Answer
Jun 12, 2018

x2+2xdx=(x+1)x2+2x212ln(x+1)+x2+2x+C

Explanation:

Complete the square:

x2+2xdx=x2+2x+11dx=(x+1)21dx

Substitute x+1=sect, dx=secttantdt, with t(0,π2):

x2+2xdx=sec2t1secttantdt

Using the trigonometric identity sec2t1=tan2t and as tant>0 for t(0,π2):

x2+2xdx=tan2tsecttantdt:

x2+2xdx=secttan2tdt

x2+2xdx=sect(sec21)dt

x2+2xdx=sec3tdtsectdt

Now the first integral is:

sectdt=sectsect+tantsect+tantdt

sectdt=sec2t+secttantsect+tantdt

sectdt=d(sect+tant)sect+tantdt

sectdt=ln|sect+tant|+c

and for the second we can integrate by parts

sec3tdt=sectsec2tdt=sectddt(tant)dt

sec3tdt=secttanttantddt(sect)dt

sec3tdt=secttantsecttan2tdt

sec3tdt=secttantsect(sec21)dt

sec3tdt=secttantsec3tdt+sectdt

The integral appears on both sides of the equation and we can solve for it:

sec3tdt=secttant2+12sectdt

Putting it together:

x2+2xdx=secttant212ln|sect+tant|+C

To undo the substitution:

sect=x+1

tant=sec21=(x+1)21=x2+2x

Finally:

x2+2xdx=(x+1)x2+2x212ln(x+1)+x2+2x+C