How do you integrate #int xsqrt(3x+ 1) dx#?

1 Answer
Apr 15, 2017

#int xsqrt(3x + 1) = 2/9x(3x + 1)^(3/2) - 4/135(3x + 1)^(5/2) + C#

Explanation:

You will need substitution in addition to integration by parts. Let #u = x# and #dv = sqrt(3x +1)#. By the power rule, #du = dx#. But we will need to make a substitution to find #v#. Let #t= 3x+ 1#. Then #dt = 3dx# and #dx = (dt)/3#.

#intsqrt(3x + 1) = sqrt(t) * (dt)/3 = 1/3sqrt(t)dt = 2/9t^(3/2) = 2/9(3x + 1)^(3/2)#

Apply integration by parts now.

#int udv = uv - int vdu#

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

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

Now let #n = 3x + 1#. Then #dn =3dx# and #dx = (dn)/3#.

#int (3x + 1)^(3/2) = 1/3int n^(3/2) dn = 2/15n^(5/2) = 2/15(3x + 1)^(5/2)#

#int xsqrt(3x + 1) = 2/9x(3x + 1)^(3/2) - 2/9(2/15)(3x +1)^(5/2) + C#

#int xsqrt(3x + 1) = 2/9x(3x + 1)^(3/2) - 4/135(3x + 1)^(5/2) + C#

Hopefully this helps!