How do I find the integral of sqrt (x^2+2x+1)?

Answer by  Taras (7)

Note, that x^2 + 2x + 1 = (x + 1)^2, so your function f(x) = sqrt(x^2+2x+1) = abs(x+1), i.e. x >= -1, f(x) = x+1 and integral f(x) = (x^2)/2 + x + C x < -1 , f(x) = -x-1 and integral f(x) = - (x^2)/2 - x + C

Answer by  VB (361)

It is sometimes helpful to rewrite the integrand. For this one, factor what is under the square root. Then simplify with the square root. If correct, it should be easy.

Answer by  blue25 (346)

(x+1)(x+1) what times what gave you x^2 and +1? to get x^2 you have to multiply two x (x times x = x^2) and one times one =1 (since both negative and positive gave you the same answer you have to check the middle one) to get 2x it have to be positive therefore it have to be two +1

Answer by  VB (361)

This one requires rewriting the expression before you can integrate. If you were to factor what is under the square root symbol, you should get (x+1)^2. The square root and the square undo each other. So you just need to integrate x+1.

Answer by  amswplusone (652)

First note that x^2+2x+1=(x+1)^2, so sqrt(that) is x+1. Thus you're integrating x+1; raise the power of x by one and divide by the new exponent, ditto for 1, making it (1/2)x^2+x. Don't forget to add a constant! So you end up with (1/2)x^2+x+C, with C an arbitrary constant.

Answer by  anrimala36 (208)

Given c*x^n where c and n are numbers and x is a variable, you find the integral by writing c/(n+1)*x^(n+1). The integral of x^2 is 1/(2+1)*x^3; the integral of 2x is 2/(1+1)*x^2; the integral of 1, or 1*x^0, is 1/(0+1)*x. Now add them. Answer = 1/3*x^3+x^2+x

Answer by  willard (874)

Integration is a linear operation, so that integral of x^2+2x +1 = integral of x^2 + 2* intregral of x + integral of 1. Just calculate each of those three integrals and then add up the answers and you are done. This works for any polynomial in x, or even any sum of other functions.