int3egral , polinomials

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January 26th 2013, 07:25 AM
Boo

int3egral , polinomials

Dear All!
please, can someone help me solve this:

$\int \frac{x^2+4}{x^2+16}dx$

would it be proper:

$}int \frac{x^2+16-12}{x^2+16}=\int \frac{x^2+16}{x^2+16dx}-\int \frac{12}{x^2+16}dx=\int(1)dx-12\int \frac{1}{x^2+16}$

$=x-12 \frac{1}{4}arctan(x/4)+C$
what to do now?
Thanks in advance!
January 26th 2013, 07:38 AM
a tutor

Re: int3egral , polinomials

It's OK to look up (or remember) a standard integral like this.

If you want to do it yourself you can use the substitution $x=4\tan\theta$

Haha. I think you finished the problem while I was typing my post.

Apart from 12 x 1/4 = 3 of course, but really you're finished.
January 26th 2013, 08:05 AM
Boo

Re: int3egral , polinomials

Thanks tutor!
Still have to check if they ment sth else!!!
:)
January 30th 2013, 05:23 AM
Boo

Re: int3egral , polinomials

Hello!
Please, the answer should be:
$\frac{\sqrt{2}}{4}arctan\frac{x-\sqrt{2}}{\sqrt{2}}+\frac{\sqrt{2}}{4}arctan\frac{ x+\sqrt{2}}{\sqrt{2}}+C$

Please, the hell, how to get it???
many thanks!!!
January 30th 2013, 05:49 AM
a tutor

Re: int3egral , polinomials

Quote:

Originally Posted by Boo

Hello!
Please, the answer should be:
$\frac{\sqrt{2}}{4}arctan\frac{x-\sqrt{2}}{\sqrt{2}}+\frac{\sqrt{2}}{4}arctan\frac{ x+\sqrt{2}}{\sqrt{2}}+C$

Please, the hell, how to get it???
many thanks!!!

No. The answer you gave in your first post in this thread is correct.

$x-3\tan^{-1}\left(\frac{x}{4}\right)+C$ .
January 30th 2013, 06:32 AM
hollywood

Re: int3egral , polinomials

The result you give is $\int \frac{x^2+4}{x^4+16} \,dx$ instead of $\int \frac{x^2+4}{x^2+16} \,dx$ . Did you copy the question correctly in your original post?

- Hollywood
January 31st 2013, 05:10 AM
Boo

Re: int3egral , polinomials

Ujoj!!!
The first one with $x^4$ is right!!!
Seems I could solve "my version" but I cant solve the one they worte in the book...
January 31st 2013, 05:13 AM
Boo

Re: int3egral , polinomials

$\int \frac{x^2+4}{x^4+^16= }$
January 31st 2013, 07:33 AM
hollywood

Re: int3egral , polinomials

You would factor $x^4+16$ as $(x^2+2\sqrt{2}x+4)(x^2-2\sqrt{2}x+4)$ and use partial fractions. It'll take a lot of work, but there's no trick or anything. Just follow the method and the answer should come right out.

- Hollywood
January 31st 2013, 08:29 AM
Boo

Re: int3egral , polinomials

$\frac{Ax+B}{x^2-2\sqrt{2}x+4}+\frac{Cx+D}{x^2-2\sqrt{2}x+4}$
$(Ax+B)(x^2+2\sqrt{2}x+4)+(Cx+D)(x^2-2\sqrt{2}x+4)=x^2+4$
$Ax^3+Bx^2+2\sqrt{2}Ax^2+2\sqrt{2}Bx+4Ax+4B+Bx^3+Cx ^2+2\sqrt{2}Bx^2+2\sqrt{2}Cx+4Bx+4C=x^2+4$

Do we eliminate now x^3 and x^1 as it should be zero-I suppose?

Am I right till now???
Many thanks!!!
February 1st 2013, 10:50 PM
hollywood

Re: int3egral , polinomials

That's the correct idea, but you're making mistakes. Here's what I got:

$\frac{x^2+4}{x^4+16}=\frac{Ax+B}{x^2-2\sqrt{2}x+4}+\frac{Cx+D}{x^2+2\sqrt{2}x+4}$

$(Ax+B)(x^2+2\sqrt{2}x+4)+(Cx+D)(x^2-2\sqrt{2}x+4)=x^2+4$

$(A+C)x^3 + (2\sqrt{2}A - 2\sqrt{2}C + B + D)x^2 +$

$(2\sqrt{2}B - 2\sqrt{2}D + 4A + 4C)x + (4B+4D) = x^2+4$

We can equate all four coefficients:
$A+C=0$
$2\sqrt{2}A - 2\sqrt{2}C + B + D=1$
$2\sqrt{2}B - 2\sqrt{2}D + 4A + 4C=0$
$4B+4D=4$

Next you can solve for A, B, C, and D - four equations in four unknowns. You should get $A=C=0$ and $B=D=\frac{1}{2}$ . Now you can integrate the two functions - do you know how?

- Hollywood