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

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Homework Help Overview

The problem involves evaluating the integral \(\int\frac{xdx}{3+\sqrt{x}}\), with a provided answer for verification. The discussion centers around the steps taken to simplify and compute the integral, including substitutions and algebraic manipulations.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts a substitution method with \(u=\sqrt{x}\) and transforms the integral accordingly. Some participants question the accuracy of the calculations and suggest checking for arithmetic errors. Others note the arbitrary nature of the constant of integration, indicating that it can account for discrepancies in constant terms.

Discussion Status

The discussion has progressed with participants providing feedback on the original poster's calculations. There is acknowledgment of a mistake in the arithmetic, and the original poster seems to have arrived at the correct form of the answer after revisiting the calculations. However, no explicit consensus is reached regarding the initial steps.

Contextual Notes

Participants are working within the constraints of homework guidelines, focusing on understanding the integral's evaluation rather than providing direct solutions. The discussion reflects an ongoing exploration of the problem rather than a definitive resolution.

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Homework Statement



[tex]\int\frac{xdx}{3+\sqrt{x}}[/tex]


Homework Equations



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

The Attempt at a Solution



[tex]u=\sqrt{x}[/tex]

[tex]u^2=x[/tex]

[tex]2udu=dx[/tex]

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

[tex]w=3+u[/tex]

[tex]w-3=u[/tex]

[tex]dw=du[/tex]

[tex]=2\int\frac{(w-3)^3dw}{w}[/tex]

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

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

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

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

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


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

I multiplied this out but terms didn't cancel. Any suggestions?
 
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looks like you got everything right, I am sure you made some calculation mistake.

what's your final answer ?
 
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:

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

[tex]=\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[/tex]

[tex]=\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[/tex]

[tex]=\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[/tex]

[tex]=\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[/tex]


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

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

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

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

Thank you for the help!
 

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