How do you integrate #int xsqrt(x+2)dx#?
1 Answer
Nov 9, 2016
# int xsqrt(x+2) dx = {2(x+2)^(3/2)(3x-4)}/15 + C #
Explanation:
Let
We can integrate this easily by substitution:
Let
# u = x+2 => (du)/dx=1 # , or# int ... du = int ... dx #
Applying the substitution we have:
# I = int (u-2)sqrtu du #
# :. I = int (u-2)u^(1/2) du #
# :. I = int u^(3/2) - 2u^(1/2) du #
# :. I = u^(5/2)/(5/2) - 2u^(3/2)/(3/2) + C #
# :. I = 2/5u^(5/2) - 4/3u^(3/2) + C #
# :. I = 2/5(x+2)^(5/2) - 4/3(x+2)^(3/2) + C #
We can put simplify by putting over a common denominator and factoring out
# :.I = (6(x+2)^(5/2) - 20(x+2)^(3/2))/15 + C #
# :.I = (x+2)^(3/2){(6(x+2) - 20)}/15 + C #
# :.I = (x+2)^(3/2){(6x+12 - 20)}/15 + C #
# :.I = (x+2)^(3/2){(6x-8)}/15 + C #
# :.I = {2(x+2)^(3/2)(3x-4)}/15 + C #
