How do you integrate #int x/sqrt(2+3x)# by parts?

2 Answers
Nov 17, 2016

# int x/sqrt(2+3x)dx = 2/27sqrt(2+3x)(3x - 4) + C#

Explanation:

You would not integrate this integral By Parts, instead you would use a substitution ;

Let # u = 2+3x => x=1/3(u-2) #
And, # (du)/dx = 3 => 1/3int ...du = int ... dx #

Applying this substitution to the problem gives us:

# int x/sqrt(2+3x)dx = int (1/3(u-2))/sqrt(u)(1/3)du#
# :. int x/sqrt(2+3x)dx = 1/9 int (u-2)u^(-1/2)du#
# :. int x/sqrt(2+3x)dx = 1/9 int u^(1/2)-2u^(-1/2)du#
# :. int x/sqrt(2+3x)dx = 1/9 {u^(3/2)/(3/2) - (2u^(1/2))/(1/2)} + C#
# :. int x/sqrt(2+3x)dx = 1/9 u^(1/2){2/3u - 4} + C#
# :. int x/sqrt(2+3x)dx = 1/9 u^(1/2){2/3(u - 6)} + C#
# :. int x/sqrt(2+3x)dx = 2/27 u^(1/2)(u - 6) + C#

Substituting for #u# gives is;

# int x/sqrt(2+3x)dx = 2/27sqrt(2+3x)(2+3x - 6) + C#
# :. int x/sqrt(2+3x)dx = 2/27sqrt(2+3x)(3x - 4) + C#

Please see below.

Explanation:

#I = int x(2+3x)^(-1/2) dx#

Let #u = x# and #dv = (2+3x)^(-1/2) dx#

This make #du = dx# and

#v = int (2+3x)^(-1/2) dx = 2/3 (2+3x)^(1/2)# (by substitution)

#I = uv-int v du#

# = 2/9x(2+3x)^(1/2) - 2/3 int (2+3x)^(1/2) dx#

# = 2/3x(2+3x)^(1/2) - 2/3 [ 2/9(2+3x)^(3/2) ] +C# (by substitution)

Clean up the algebra as desired.