# Integral Problem

## 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?

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

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.

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!