How do you integrate #int xsqrt(3x+ 1) dx#?
1 Answer
Explanation:
You will need substitution in addition to integration by parts. Let
#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
#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!