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

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In summary, the homework statement is that $\int\frac{xdx}{3+\sqrt{x}}eq 2\int\frac{(u^3)du}{3+u}w=3+uw-3=udw=duu=\sqrt{x}u^2=x2udu=dx\int\frac{xdx}{3+\sqrt{x}} = 2\int\frac{(u^3)du}{3+u}w=3+uw-3=udw=duu=\sq
  • #1
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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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  • #2
looks like you got everything right, I am sure you made some calculation mistake.

what's your final answer ?
 
  • #3
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.
 
  • #4
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!
 

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

What does the integral \int\frac{xdx}{3+\sqrt{x}} represent?

The integral \int\frac{xdx}{3+\sqrt{x}} represents the area under the curve of the function \frac{x}{3+\sqrt{x}} with respect to x. It is a mathematical operation used to calculate the total area of a specific region on a graph.

How do I solve the integral \int\frac{xdx}{3+\sqrt{x}}?

To solve this integral, you can use the substitution method. Let u = 3+\sqrt{x}, then du = \frac{1}{2\sqrt{x}}dx. Substituting these values into the original integral, we get \int\frac{xdx}{3+\sqrt{x}} = \int\frac{2u-6}{u}du. This can be solved by using basic integration rules.

What is the domain and range of the function \frac{x}{3+\sqrt{x}}?

The domain of the function \frac{x}{3+\sqrt{x}} is all real numbers greater than or equal to 0, since the square root of a negative number is undefined. The range of the function is all real numbers greater than or equal to 0, as the function approaches 0 as x approaches infinity.

What is the significance of the constant 3 in the denominator of the function \frac{x}{3+\sqrt{x}}?

The constant 3 represents a vertical shift of the function. It determines the position of the graph on the y-axis. Without this constant, the graph would pass through the origin and have a different shape.

How can I use the integral \int\frac{xdx}{3+\sqrt{x}} to solve real-world problems?

The integral \int\frac{xdx}{3+\sqrt{x}} can be used to solve various real-world problems involving rates of change, such as finding the total amount of a substance in a given time period or calculating the area under a velocity-time graph to determine displacement.

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