# Math Help - integral ...

1. ## integral ...

What is the problem here I get diffrent result ....

$\int_0^2 \frac{\frac{5}{2}}{4x^4+25}= \frac{5}{2}\int_0^2 \frac{1}{(2x^2)^2+5^2}dx = \frac{5*1}{2*5} arctan (\frac{2x^2}{5})$

2. Originally Posted by gilyos
What is the problem here I get diffrent result ....

$\int_0^2 \frac{\frac{5}{2}}{4x^4+25}= \frac{5}{2}\int_0^2 {1}{(2x^2)^2+5^2}dx = \frac{5*1}{2*5} arctan (\frac{2x^2}{5}$
I think you mean $\frac{1}{(2x^2)^2+5^2}$ not ${1}{(2x^2)^2+5^2}$

Look to this formula carefaully (Assuming a dont equal 0):
$\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} arctan \frac{x}{a} + C$

Are you sure your problem is similar to it?
i.e. when you substituting $u=2x^2$ , You will get your du in the numerator?

You forgot the $dx$ in the first integral. Also its a definite integral, Its result must be a number not a function.

3. I check and the result before the answer and afther the answer is different ... why ?

4. I told you : you have a definite integral, its result must be a number not a function.

Did you substitute $u=2x^2$ ?
Try it and you will see what I mean.

5. You can use partial fractions:
$4x^4+25 = (2x^2)^2+20x^2+5^2-20x^2 = (2x^2+5)^2-(\sqrt{20}x)^2 =$
$= (2x^2-\sqrt{20}x+5)(2x^2+\sqrt{20}x+5)$

6. Originally Posted by gilyos
What is the problem here I get diffrent result ....

$\int_0^2 \frac{\frac{5}{2}}{4x^4+25}= \frac{5}{2}\int_0^2 \frac{1}{(2x^2)^2+5^2}dx = \frac{5*1}{2*5} arctan (\frac{2x^2}{5})$
Originally Posted by Unbeatable0
You can use partial fractions:
$4x^4+25 = (2x^2)^2+20x^2+5^2-20x^2 = (2x^2+5)^2-(\sqrt{20}x)^2 =$
$= (2x^2-\sqrt{20}x+5)(2x^2+\sqrt{20}x+5)$
Strange. This seems to be exactly what I suggested to you last time you posted this question.

7. Originally Posted by Drexel28
Strange. This seems to be exactly what I suggested to you last time you posted this question.
Oh sorry I didn't notice. I found it now here
http://www.mathhelpforum.com/math-he...rp-8-11-a.html

8. Originally Posted by Unbeatable0
Oh sorry I didn't notice. I found it now here
http://www.mathhelpforum.com/math-he...rp-8-11-a.html
Not at all your fault of course!