# 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 #