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 $\displaystyle x=4\tan\theta$

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.

Re: int3egral , polinomials

Thanks tutor!

Still have to check if they ment sth else!!!

:)

Re: int3egral , polinomials

Hello!

Please, the answer should be:

$\displaystyle \frac{\sqrt{2}}{4}arctan\frac{x-\sqrt{2}}{\sqrt{2}}+\frac{\sqrt{2}}{4}arctan\frac{ x+\sqrt{2}}{\sqrt{2}}+C$

Please, the hell, how to get it???

many thanks!!!

Re: int3egral , polinomials

Quote:

Originally Posted by

**Boo** Hello!

Please, the answer should be:

$\displaystyle \frac{\sqrt{2}}{4}arctan\frac{x-\sqrt{2}}{\sqrt{2}}+\frac{\sqrt{2}}{4}arctan\frac{ x+\sqrt{2}}{\sqrt{2}}+C$

Please, the hell, how to get it???

many thanks!!!

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

$\displaystyle x-3\tan^{-1}\left(\frac{x}{4}\right)+C$.

Re: int3egral , polinomials

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

- Hollywood

Re: int3egral , polinomials

Ujoj!!!

The first one with $\displaystyle x^4$is right!!!

Seems I could solve "my version" but I cant solve the one they worte in the book...

Re: int3egral , polinomials

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

Re: int3egral , polinomials

You would factor $\displaystyle x^4+16$ as $\displaystyle (x^2+2\sqrt{2}x+4)(x^2-2\sqrt{2}x+4)$ 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

Re: int3egral , polinomials

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

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

$\displaystyle Ax^3+Bx^2+2\sqrt{2}Ax^2+2\sqrt{2}Bx+4Ax+4B+Bx^3+Cx ^2+2\sqrt{2}Bx^2+2\sqrt{2}Cx+4Bx+4C=x^2+4$

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

Am I right till now???

Many thanks!!!

Re: int3egral , polinomials

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

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

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

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

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

We can equate all four coefficients:

$\displaystyle A+C=0$

$\displaystyle 2\sqrt{2}A - 2\sqrt{2}C + B + D=1$

$\displaystyle 2\sqrt{2}B - 2\sqrt{2}D + 4A + 4C=0$

$\displaystyle 4B+4D=4$

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

- Hollywood