Solve Integral Problem: \int\frac{xdx}{3+\sqrt{x}}

[3.141592654]

Homework Statement

\int\frac{xdx}{3+\sqrt{x}}

Homework Equations

The answer is given: \frac{2}{3}x^\frac{3}{2}-3x+18\sqrt{x}-54ln(3+\sqrt{x})+C

The Attempt at a Solution

u=\sqrt{x}

u^2=x

2udu=dx

\int\frac{xdx}{3+\sqrt{x}} = 2\int\frac{(u^3)du}{3+u}

w=3+u

w-3=u

dw=du

=2\int\frac{(w-3)^3dw}{w}

=2\int\frac{(w^3-9w^2+27w-27)dw}{w}

=2\int\((w^2-9w+27-\frac{27}{w})dw

=2\int\(w^2dw-18\int\(wdw+54\int\(dw-54\int\frac{dw}{w}

=2\frac{w^3}{3}-18\frac{w^2}{2}+54w-54ln|w|+C

=\frac{2}{3}(3+u)^3-9(3+u)^2+54(3+u)-54ln|3+u|+C


=\frac{2}{3}(3+\sqrt{x})^3-9(3+\sqrt{x})^2+54(3+\sqrt{x})-54ln|3+\sqrt{x}|+C

I multiplied this out but terms didn't cancel. Any suggestions?

 

[Roni1985]

looks like you got everything right, I am sure you made some calculation mistake.

what's your final answer ?

 

[Mathnerdmo]

If you multiply out those terms, you'll get the given answer, but with an additional term of + 99. (or something of that sort)

Assuming this is your problem, you just need to remember that the constant of integration C is arbitrary, so it can "absorb" any constant terms.

 

[3.141592654]

Alright it looks like I just made a mistake last time:

=\frac{2}{3}(3+\sqrt{x})^3-9(3+\sqrt{x})^2+54(3+\sqrt{x})-54ln|3+\sqrt{x}|+C

=\frac{2}{3}(3+\sqrt{x})(3+\sqrt{x})(3+\sqrt{x})-9(3+\sqrt{x})(3+\sqrt{x})+54(3+\sqrt{x})-54ln|3+\sqrt{x}|+C

=\frac{2}{3}(9+6\sqrt{x}+x)(3+\sqrt{x})-9(9+6\sqrt{x}+x)+54(3+\sqrt{x})-54ln|3+\sqrt{x}|+C

=\frac{2}{3}(27+9\sqrt{x}+18\sqrt{x}+6x+3x+x^\frac{3}{2})-81-54\sqrt{x}-9x+162+54\sqrt{x}-54ln|3+\sqrt{x}|+C

=\frac{2}{3}(27+9\sqrt{x}+18\sqrt{x}+6x+3x+x^\frac{3}{2})-81+162-54\sqrt{x}+54\sqrt{x}-9x-54ln|3+\sqrt{x}|+C


=\frac{2}{3}(27+27\sqrt{x}+9x+x^\frac{3}{2})+81-9x-54ln|3+\sqrt{x}|+C

=18+18\sqrt{x}+6x+\frac{2}{3}x^\frac{3}{2}+81-9x-54ln|3+\sqrt{x}|+C

=99+18\sqrt{x}-3x+\frac{2}{3}x^\frac{3}{2}-54ln|3+\sqrt{x}|+C

=\frac{2}{3}x^\frac{3}{2}-3x+18\sqrt{x}-54ln|3+\sqrt{x}|+C

Thank you for the help!