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

## Homework Statement

$\int\frac{dx}{x^{2}*\sqrt{6x+1}}$

## The Attempt at a Solution

t=sqrt(6x+1)
so x=t$^{2}$-1
dx=2t dt

so new integral is

2*$\int\frac{t}{(t^{2}-1)^{2}*t}$

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

By partial fractions I get the new integral to be

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

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|$\sqrt{6x+1}$+1|)-(ln|$\sqrt{6x+1}$-1|)-$\frac{1}{\sqrt{6x+1}+1}$-$\frac{1}{\sqrt{6x+1}-1}$)

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?

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SammyS
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## Homework Statement

$\int\frac{dx}{x^{2}*\sqrt{6x+1}}$

## The Attempt at a Solution

t=sqrt(6x+1)
so x=t$^{2}$-1
dx=2t dt

so new integral is

2*$\int\frac{t}{(t^{2}-1)^{2}*t}$

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

By partial fractions I get the new integral to be

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

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|$\sqrt{6x+1}$+1|)-(ln|$\sqrt{6x+1}$-1|)-$\frac{1}{\sqrt{6x+1}+1}$-$\frac{1}{\sqrt{6x+1}-1}$)

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
$\displaystyle t=\sqrt{6x+1}$​
then
$\displaystyle x=\frac{t^2-1}{6}$​

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