Integral of 1/[(x^2)*(sqrt(6x+1))]

  • Thread starter skyturnred
  • Start date
  • Tags
    Integral
In summary, the conversation discusses solving an integral involving x, t, and a constant of 6. By using partial fractions and substitutions, the integral can be simplified to a final answer involving natural logarithms and the original constant. The speaker also expresses frustration with making a mistake in the process and asks for tips on avoiding similar mistakes in the future.
  • #1
skyturnred
118
0

Homework Statement



[itex]\int\frac{dx}{x^{2}*\sqrt{6x+1}}[/itex]

Homework Equations


The Attempt at a Solution



t=sqrt(6x+1)
so x=t[itex]^{2}[/itex]-1
dx=2t dt

so new integral is

2*[itex]\int\frac{t}{(t^{2}-1)^{2}*t}[/itex]

the t's cancel and then i can turn the denominator into (t+1)[itex]^{2}[/itex]*(t-1)[itex]^{2}[/itex]

By partial fractions I get the new integral to be

(1/4)*( [itex]\int\frac{dt}{t+1}[/itex]+[itex]\int\frac{dt}{(t+1)^{2}}[/itex]-[itex]\int\frac{dt}{(t-1)}[/itex]+[itex]\int\frac{dt}{(t-1)^{2}}[/itex] )

so the first and third are easy. The second I make a substitution equal to t+1. The 4th I make a substitution equal to t-1.

After finding all the integrals I replace all the 't's with (sqrt(6x+1)) and my final answer is(1/4)*((ln|[itex]\sqrt{6x+1}[/itex]+1|)-(ln|[itex]\sqrt{6x+1}[/itex]-1|)-[itex]\frac{1}{\sqrt{6x+1}+1}[/itex]-[itex]\frac{1}{\sqrt{6x+1}-1}[/itex])

But apparently it's wrong and after checking over it I can't find out where. I am getting super frustrated because even when I KNOW the procedure for a problem, I am more often wrong than right. Does anyone also have any general tips to prevent this?
 
Physics news on Phys.org
  • #2
skyturnred said:

Homework Statement



[itex]\int\frac{dx}{x^{2}*\sqrt{6x+1}}[/itex]

Homework Equations





The Attempt at a Solution



t=sqrt(6x+1)
so x=t[itex]^{2}[/itex]-1
dx=2t dt

so new integral is

2*[itex]\int\frac{t}{(t^{2}-1)^{2}*t}[/itex]

the t's cancel and then i can turn the denominator into (t+1)[itex]^{2}[/itex]*(t-1)[itex]^{2}[/itex]

By partial fractions I get the new integral to be

(1/4)*( [itex]\int\frac{dt}{t+1}[/itex]+[itex]\int\frac{dt}{(t+1)^{2}}[/itex]-[itex]\int\frac{dt}{(t-1)}[/itex]+[itex]\int\frac{dt}{(t-1)^{2}}[/itex] )

so the first and third are easy. The second I make a substitution equal to t+1. The 4th I make a substitution equal to t-1.

After finding all the integrals I replace all the 't's with (sqrt(6x+1)) and my final answer is


(1/4)*((ln|[itex]\sqrt{6x+1}[/itex]+1|)-(ln|[itex]\sqrt{6x+1}[/itex]-1|)-[itex]\frac{1}{\sqrt{6x+1}+1}[/itex]-[itex]\frac{1}{\sqrt{6x+1}-1}[/itex])

But apparently it's wrong and after checking over it I can't find out where. I am getting super frustrated because even when I KNOW the procedure for a problem, I am more often wrong than right. Does anyone also have any general tips to prevent this?
I haven't checked the whole thing yet, but if
[itex]\displaystyle t=\sqrt{6x+1}[/itex]​
then
[itex]\displaystyle x=\frac{t^2-1}{6}[/itex]​
 
  • #3
OK thanks. That's incredibly frustrating that I made such a dumb mistake but at least I know now
 

1. What is the integral of 1/[(x^2)*(sqrt(6x+1))]?

The integral of 1/[(x^2)*(sqrt(6x+1))] is equal to 1/(3*sqrt(6x+1))+C, where C is the constant of integration.

2. How do you evaluate the integral of 1/[(x^2)*(sqrt(6x+1))]?

To evaluate the integral of 1/[(x^2)*(sqrt(6x+1))], you can use the substitution method by letting u = 6x+1. After solving for u, you can then substitute back into the original integral and integrate using the power rule.

3. Can the integral of 1/[(x^2)*(sqrt(6x+1))] be simplified further?

No, the integral of 1/[(x^2)*(sqrt(6x+1))] cannot be simplified further. However, you can rewrite it as 1/[(3x)*(sqrt(6x+1))] if you prefer.

4. Are there any special cases to consider when solving the integral of 1/[(x^2)*(sqrt(6x+1))]?

Yes, when x = 0, the integral is undefined. Additionally, if x = -1/6, the integral becomes an arctangent function.

5. How can the integral of 1/[(x^2)*(sqrt(6x+1))] be applied in real-world situations?

The integral of 1/[(x^2)*(sqrt(6x+1))] can be used in various fields such as physics and engineering to calculate the work done by a variable force or the acceleration due to gravity in a non-uniform gravitational field. It can also be used in economics to calculate marginal cost in production processes.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
754
  • Calculus and Beyond Homework Help
Replies
22
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
715
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
558
  • Calculus and Beyond Homework Help
Replies
10
Views
416
  • Calculus and Beyond Homework Help
Replies
3
Views
309
  • Calculus and Beyond Homework Help
Replies
4
Views
105
  • Calculus and Beyond Homework Help
Replies
2
Views
135
  • Calculus and Beyond Homework Help
Replies
16
Views
551
Back
Top