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

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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?
 

Answers and Replies

  • #2
SammyS
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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
118
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OK thanks. That's incredibly frustrating that I made such a dumb mistake but at least I know now
 

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