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

[skyturnred]

Homework Statement

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

Homework Equations

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?

 

[SammyS]

Homework Statement

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

Homework Equations

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}​

 

[skyturnred]

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