# Thread: Integration of a Fraction

1. ## Integration of a Fraction

I was wondering if there was any way of me integrating this by using a normal substitution technique rather than subsitution using trigonometric values.

$\int \frac{1}{x\sqrt{x^2 -1}}$

I was thinking along the lines of having $u = \sqrt{x^2 -1}$

I tried it out this way, but it didnt seem to work, of it does, could someone maybe explain how. Thanks in advance

2. Originally Posted by brd_7
I was wondering if there was any way of me integrating this by using a normal substitution technique rather than subsitution using trigonometric values.

$\int \frac{1}{x\sqrt{x^2 -1}}$

I was thinking along the lines of having $u = \sqrt{x^2 -1}$

I tried it out this way, but it didnt seem to work, of it does, could someone maybe explain how. Thanks in advance
you can make this work (i would have probably tried a trig substitution, but your way is fine, probably easier).

$\int \frac 1{x \sqrt{x^2 - 1}}~dx$

note that this is the same as: $\int \frac x{x^2 \sqrt{x^2 - 1}}~dx$ you'll see why i did that in a minute.

Let $u = \sqrt{x^2 - 1} \implies x^2 = u^2 + 1$

$\Rightarrow du = \frac {x}{\sqrt{x^2 - 1}}~dx$

So, our integral becomes

$\int \frac 1{\left( u^2 + 1 \right)u}~du$

by partial fractions we find that this is:

$\int \left(\frac 1u - \frac u{u^2 + 1} \right)~du$

and i think you can take it from there, right?

(by the way, there's actually a formula for this integral, look it up in your text book. it will be an equivalent answer. i think it will be arcsec(x) or something like that)

3. Originally Posted by Jhevon
you can make this work (i would have probably tried a trig substitution, but your way is fine, probably easier).

$\int \frac 1{x \sqrt{x^2 - 1}}~dx$

note that this is the same as: $\int \frac x{x^2 \sqrt{x^2 - 1}}~dx$ you'll see why i did that in a minute.

Let $u = \sqrt{x^2 - 1} \implies x^2 = u^2 + 1$

$\Rightarrow du = \frac {x}{\sqrt{x^2 - 1}}~dx$

So, our integral becomes

$\int \frac 1{\left( u^2 + 1 \right)u}~du$

by partial fractions we find that this is:

$\int \left(\frac 1u - \frac u{u^2 + 1} \right)~du$

and i think you can take it from there, right?
Ok Thank you so much, im going to try and show all the steps properly now and then complete it myself.. I shall post you with an update. Thanks again

4. I cant figure out how you came about getting your answer when doing partial fractions.. i got 1/u.. but not the 2nd part...

5. I eventually get..

$ln(\sqrt{x^2-1}) - \frac{ln(x^2)}{2}$

6. Originally Posted by brd_7
I cant figure out how you came about getting your answer when doing partial fractions.. i got 1/u.. but not the 2nd part...
By the method of partial fractions,

$\frac 1{\left( u^2 + 1 \right) u} = \frac Au + \frac {Bu + C}{u^2 + 1}$

multiply through by $u \left( u^2 + 1 \right)$, we get:

$1 = A \left( u^2 + 1 \right) + (Bu + C) u$

$\Rightarrow 1 = (A + B)u^2 + Cu + A$

equating coefficients of like powers of $u$, we obtain the system:

$A + B = 0$ ........(1)
$C = 0$ ..............(2)
$A = 1$ ..............(3)

plug (3) into (1), we get $B = -1$

Thus, $\frac 1{\left( u^2 + 1 \right) u} = \frac 1u - \frac u{u^2 + 1}$

7. Originally Posted by brd_7
I eventually get..

$ln(\sqrt{x^2-1}) - \frac{ln(x^2)}{2}$
correct. you could simplify that, but it's fine. don't forget your arbitrary constant

8. The constant wasnt nescessary for this question, its part of a series question and my integral values are t and 1 and i simply have to show if this converges or diverges. 23 questions later, my work is complete. Thank you!

9. Originally Posted by brd_7
$\int \frac{1}{x\sqrt{x^2 -1}}$

I was thinking along the lines of having $u = \sqrt{x^2 -1}$
Mmm. Okay, tryin' that substitution yields

$u = \sqrt {x^2 - 1} \implies du = \frac{x}
{{\sqrt {x^2 - 1} }}\,dx.$

The integral becomes to

$\int\frac1{u^2+1}\,du=\arctan u+k=\arctan\left(\sqrt {x^2 - 1}\right)+k.$

--

About this $\frac{1}
{{u\left( {u^2 + 1} \right)}},$
try to avoid partial fractions:

$\frac{1}
{{u\left( {u^2 + 1} \right)}} = \frac{{u^2 + 1 - u^2 }}
{{u\left( {u^2 + 1} \right)}} = \frac{1}
{u} - \frac{u}
{{u^2 + 1}}.$