# i need help in integration question?

• Jan 24th 2010, 10:25 AM
khhratmath123
i need help in integration question?
solve this question.......
• Jan 24th 2010, 10:26 AM
craig
Quote:

Originally Posted by khhratmath123
solve this question.......

Show us your work so far and then we can guide you through where you're stuck.

As a pointer, the way I would approach this is to factorise the denominator, and then use partial fractions to get two fractions which are easy to integrate.
• Jan 24th 2010, 10:42 AM
craig
Quote:

Originally Posted by khhratmath123
solve this question.......

Also, if you are having problems factorising the fraction, remember you can take a factor of -1 out, so your problem becomes:

$(-1)\int^{-2}_{-3}\frac{1}{x^2 + 6x+5}dx$
• Jan 24th 2010, 10:58 AM
khhratmath123
i have solved this question but i have doubt....plz check..is it right?
• Jan 24th 2010, 11:12 AM
Pulock2009
this was easy:

∫((dx)/(x²+6x+5))
=∫((dx)/((x+1)(x+5)))

now use partial fractions ,get the constants and put the limits.

your answer is definitely wrong as it should come in terms of logs.
• Jan 24th 2010, 11:15 AM
VonNemo19
Quote:

Originally Posted by khhratmath123
i have solved this question but i have doubt....plz check..is it right?

Correct.
• Jan 24th 2010, 11:16 AM
craig
Quote:

Originally Posted by khhratmath123
i have solved this question but i have doubt....plz check..is it right?

No. I see what you have done, but your answer should be:

$[\frac{1}{2}tanh^{-1}(\frac{x+3}{2})]^{-2}_{-3}$

Or the way that me and Pulock suggested, this way you would get the actual logs instead of the inverse hyperbolic.
• Jan 24th 2010, 11:23 AM
VonNemo19
Quote:

Originally Posted by craig
No.

$\int\frac{dx}{(2)^2-(x+3)^2}$

Let $u=x+3\Rightarrow{du}=dx$

$\Rightarrow\int\frac{dx}{(2)^2-(x+3)^2}=\int\frac{du}{(2)^2-(u)^2}$
• Jan 24th 2010, 11:28 AM
craig
Quote:

Originally Posted by VonNemo19
$\int\frac{dx}{(2)^2-(x+3)^2}$

Let $u=x+3\Rightarrow{du}=dx$

$\Rightarrow\int\frac{dx}{(2)^2-(x+3)^2}=\int\frac{du}{(2)^2-(u)^2}$

I see how you derived that. However, if it was:

$\int\frac{du}{\sqrt{(2)^2-(u)^2}}$ that would make his answer right.

But as far as I remember, $\int\frac{du}{1-u^2} = tanh^{-1} u$ ?
• Jan 24th 2010, 11:36 AM
VonNemo19
Quote:

Originally Posted by craig
I see how you derived that. However, if it was:

$\int\frac{du}{\sqrt{(2)^2-(u)^2}}$ that would make his answer right.

But as far as I remember, $\int\frac{du}{1-u^2} = tanh^{-1} u$ ?

(Lipssealed) My mind saw a radical where there was none.
• Jan 24th 2010, 11:40 AM
craig
Quote:

Originally Posted by VonNemo19
(Lipssealed) My mind saw a radical where there was none.

Easy to do ;)
• Jan 24th 2010, 08:03 PM
khhratmath123
thanks for all
i have solved by partial fraction method