How do you evaluate the integral of #int sqrt (x^2 + 2x) dx#?

Question

Emma Aldridge · Accepted Answer

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

Complete the square:

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

Substitute #x+1 = sect#, #dx = sect tant dt#, with #t in (0,pi/2)#:

#int sqrt(x^2+2x)dx = sqrt(sec^2t-1) sect tant dt#

Using the trigonometric identity #sec^2t-1 = tan^2t# and as #tan t > 0# for #t in (0,pi/2)#:

#int sqrt(x^2+2x)dx = int sqrt(tan^2t) sect tant dt#:

#int sqrt(x^2+2x)dx = int sect tan^2t dt#

#int sqrt(x^2+2x)dx = int sect (sec^2-1)dt#

#int sqrt(x^2+2x)dx = int sec^3t dt - int sect dt #

Now the first integral is:

#int sect dt = int sect (sect+tant)/(sect+tant)dt#

#int sect dt = int (sec^2t+sect tant)/(sect+tant)dt#

#int sect dt = int (d(sect +tant))/(sect+tant)dt#

#int sect dt =ln abs (sect+tant) +c#

and for the second we can integrate by parts

#int sec^3t dt = int sect sec^2t dt = int sect d/dt(tant) dt#

#int sec^3t dt = sect tant - int tant d/dt (sect)dt#

#int sec^3t dt = sect tant - int sect tan^2t dt#

#int sec^3t dt = sect tant - int sect (sec^2-1)dt#

#int sec^3t dt = sect tant - int sec^3t dt + int sect dt#

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

#int sec^3t dt =(sect tant)/2 +1/2 int sect dt#

Putting it together:

#int sqrt(x^2+2x)dx = (sect tant)/2 - 1/2ln abs (sect+tant) +C#

To undo the substitution:

#sect = x+1#

#tant = sqrt(sec^2-1) = sqrt((x+1)^2 -1) = sqrt(x^2+2x)#

Finally:

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

Isaiah Allen · Answer

undefined To evaluate the integral of ∫√(x^2 + 2x) dx, we can use a trigonometric substitution. Let's rewrite the integrand in a form suitable for this substitution:

x^2 + 2x = (x + 1)^2 - 1

Now, let u = x + 1, so du = dx. Substitute these into the integral:

∫√(u^2 - 1) du

This integral can be evaluated using a trigonometric substitution. Let u = sec(theta), so du = sec(theta)tan(theta) d(theta). Also, recall that sec^2(theta) - 1 = tan^2(theta), so √(u^2 - 1) = tan(theta). Substitute these into the integral:

∫tan(theta) * sec(theta)tan(theta) d(theta)

= ∫sec(theta)tan^2(theta) d(theta)

= ∫sec(theta)(sec^2(theta) - 1) d(theta)

= ∫sec^3(theta) d(theta) - ∫sec(theta) d(theta)

This integral can be solved using the reduction formula for ∫sec^n(theta) d(theta):

∫sec^n(theta) d(theta) = (1/(n-1)) sec^(n-2)(theta)tan(theta) + (n-2)/(n-1) ∫sec^(n-2)(theta) d(theta)

Applying this formula with n = 3 to the first integral, we get:

(1/2)sec(theta)tan(theta) + (1/2) ∫sec(theta) d(theta)

= (1/2)sec(theta)tan(theta) + (1/2)ln|sec(theta) + tan(theta)| + C

Finally, substitute back u = sec(theta) and use the fact that sec(theta) = x + 1 and tan(theta) = √(x^2 + 2x + 1) = √((x+1)^2) = |x+1|, to get the final answer:

(1/2)(x + 1)√(x^2 + 2x + 1) + (1/2)ln|x + 1 + √(x^2 + 2x + 1)| + C