Best method to solve this integral?

In summary, the integral \int\frac{dx}{x^2\sqrt{4x+1}} can be solved using trig substitution and partial fractions. The final answer involves the natural logarithm and the inverse hyperbolic tangent function.
  • #1
G01
Homework Helper
Gold Member
2,704
19
[tex]\int\frac{dx}{x^2\sqrt{4x+1}} [/tex] Can someone give me a hint? I've been working on this forever.
 
Physics news on Phys.org
  • #2
Let [tex]u=\sqrt{4x+1} \Rightarrow du=\frac{4dx}{\sqrt{4x+1}}[/tex]
also, [tex]\frac{1}{x^2}=\frac{16}{(u^2-1)^2}[/tex]
then use trig substitution
 
  • #3
NVM, my calculation was wrong. but i can tell you that the answer involves arctanh.
 
Last edited:
  • #4
This will give

[tex]\int\frac{dx}{x^2\sqrt{4x+1}} = 4\int \frac{du}{(u^2-1)^2}[/tex]
 
  • #5
ok using trig substitution, i got the integral of cos^2x / sin^3x dx. Is this correct so far?
 
  • #6
OK i think i may have gotten this. Is this the answer.

[tex] \frac{-\sqrt{4x + 1}}{2x} - 2\ln|\frac{\sqrt{4x + 1} - 1}{\sqrt{4x}}| +C [/tex]
 
  • #7
I get

[tex]\int\frac{dx}{x^2\sqrt{4x+1}} = \log \left| \frac{\sqrt{4x+1}+1}{\sqrt{4x+1}-1}\right| -\frac{\sqrt{4x+1}}{2x}+C[/tex]

you can always differentiate to check.

EDIT: I forgot the factor of 4.
 
Last edited:
  • #8
EDIT: Oops! should be:

Let [tex]u=\sqrt{4x+1} \Rightarrow du=\frac{1}{2}\frac{4dx}{\sqrt{4x+1}}[/tex]
also, [tex]\frac{1}{x^2}=\frac{16}{(u^2-1)^2}[/tex]

to give

[tex]\int\frac{dx}{x^2\sqrt{4x+1}} = 8\int \frac{du}{(u^2-1)^2}= 2\log \left| \frac{\sqrt{4x+1}+1}{\sqrt{4x+1}-1}\right| -\frac{\sqrt{4x+1}}{x}+C[/tex]
 
Last edited:
  • #9
BTW, for real x,

[tex]\mbox{arctanh}^{-1}(x) = \frac{1}{2}\log\left| \frac{x+1}{x-1}\right|[/tex]

which would help if you went to www.integrals.com
 
  • #10
[tex]8\int \frac{du}{(u^2-1)^2}[/tex]

Im screwing up this integral somwhere. Did you use u = sec[tex]\theta[/tex]? And did you end up with:

[tex]-4\csc\theta\cot\theta - 4\ln|\csc\theta - \cot\theta| + C [/tex]

Thanks a lot for all of this help.
 
  • #11
You didn't screw-up, I did. My bad, do this rather:

[tex]\frac{8}{(u^2-1)^2}=\frac{8}{(u-1)^2(u+1)^2}[/tex]

and now partial fractions

[tex]\frac{8}{(u-1)^2(u+1)^2}=\frac{A}{u-1}+\frac{B}{(u-1)^2}+\frac{C}{u+1}+\frac{D}{(u+1)^2}[/tex]

cross-multiply to get

[tex]8=A(u-1)(u+1)^2+B(u+1)^2+C(u-1)^2(u+1)+D(u-1)^2[/tex]

plug-in u=1 to get 8=4B or B=2;

plug-in u=-1 to get 8=4D or D=2;

plug-in u=0 to get 8=-A+B+C+D, but B=D=2, so 4=-A+C

plug-in u=2 to get 8=9A+9B+3C+D, but B=D=2, so -12=9A+3C
solving these two equations gives A=-2 and C=2. Finally, we get

[tex]\int\frac{8du}{(u-1)^2(u+1)^2}=\int\left(\frac{-2}{u-1}+\frac{2}{(u-1)^2}+\frac{2}{u+1}+\frac{2}{(u+1)^2}\right) du = -2\log |u-1|-\frac{2}{u-1}+2\log |u+1|-\frac{2}{u+1}+C[/tex]
[tex]= 2\log \left| \frac{u+1}{u-1}\right|-\frac{4u}{u^2-1}+C[/tex]

but [tex]u=\sqrt{4x+1}[/tex] so...
 
  • #12
ok thanks a lot man...
 

1. What is an integral?

An integral is a mathematical concept that represents the area under a curve on a graph. It is used to find the total value of a function or the accumulation of a quantity over a given interval.

2. Why is it important to find the best method to solve an integral?

Finding the best method to solve an integral is important because it allows for the most accurate and efficient solution. Different methods may be more suitable for certain types of integrals, and using the best method can save time and effort.

3. What are the common methods used to solve an integral?

There are several common methods used to solve an integral, including substitution, integration by parts, partial fractions, and trigonometric substitution. Each method has its own advantages and is suitable for different types of integrals.

4. How do I know which method to use for a specific integral?

The best way to determine which method to use for a specific integral is to first identify the type of integral it is (e.g. trigonometric, rational, exponential). Then, you can use a chart or guide to match the type of integral with the appropriate method.

5. Can I use technology to solve integrals?

Yes, there are many online integral calculators and computer software programs that can solve integrals for you. However, it is still important to understand the different methods and concepts behind solving integrals in order to use technology effectively.

Similar threads

  • Calculus and Beyond Homework Help
Replies
22
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
657
  • Calculus and Beyond Homework Help
Replies
7
Views
641
  • Calculus and Beyond Homework Help
Replies
10
Views
288
  • Calculus and Beyond Homework Help
Replies
3
Views
269
  • Calculus and Beyond Homework Help
Replies
8
Views
720
  • Calculus and Beyond Homework Help
Replies
7
Views
643
  • Calculus and Beyond Homework Help
Replies
3
Views
538
  • Calculus and Beyond Homework Help
Replies
20
Views
385
  • Calculus and Beyond Homework Help
Replies
5
Views
741
Back
Top