# 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?

## Answers and Replies

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!