Correct method of integration?

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Will T

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Correct method of integration?

« on: November 10, 2012, 11:38:17 pm »

$\int \frac{1}{x^4+1}\,dx$

$\int \frac{1}{x^4+2x^2+1-2x^2}\,dx$

$\int \frac{1}{(x^2+1)^2-2x^2}\,dx$

$\int \frac{1}{(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)}\,dx$

$\frac{1}{(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)}=\frac{Ax+B}{(x^2-\sqrt{2}x+1)}+\frac{Cx+D}{(x^2+\sqrt{2}x+1)}$

$\frac{1}{(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)}=\frac{(Ax+B)(x^2+\sqrt{2}x+1)+(Cx+D)(x^2-\sqrt{2}x+1)}{(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)}$

$1=(Ax+B)(x^2+\sqrt{2}+1)+(Cx+D)(x^2-\sqrt{2}+1)$

$1=Ax^3+\sqrt{2}Ax^2+Bx^2+Ax+\sqrt{2}Bx+B+Cx^3-\sqrt{2}Cx^2+Dx^2+Cx-\sqrt{2}Dx+D$

$1=(A+C)x^3+(B+D+\sqrt{2}A-\sqrt{2}C)x^2+(A+C+\sqrt{2}B-\sqrt{2}D)x+(B+D)$

$\therefore A+C=0, (B+D+\sqrt{2}A-\sqrt{2}C)=0, (A+C+\sqrt{2}B-\sqrt{2}D)=0, (B+D)=1$

$\therefore A=-\frac{\sqrt{2}}{4}, B=\frac{1}{2}, C=\frac{\sqrt{2}}{4},D=\frac{1}{2}$

$\therefore \int \frac{1}{(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)}\,dx=\int \frac{2-\sqrt{2}x}{4(x^2-\sqrt{2}x+1)}+\frac{2+\sqrt{2}x}{4(x^2+\sqrt{2}x+1)}\,dx$

$\int \frac{2-\sqrt{2}x}{4(x^2-\sqrt{2}x+1)}\,dx = \int \frac{1-\sqrt{2}x}{4(x^2-\sqrt{2}x+1)}+\frac{1}{4(x^2-\sqrt{2}x+1)}\,dx$

$\int \frac{1-\sqrt{2}x}{4(x^2-\sqrt{2}x+1)}\,dx=-\frac{\sqrt{2}}{8}\int \frac{2x-\sqrt{2}}{(x^2-\sqrt{2}x+1)}\,dx=-\frac{\sqrt{2}}{8}\ln(x^2-\sqrt{2}x+1)$

$\int \frac{1}{4(x^2-\sqrt{2}x+1)}\,dx=\frac{\sqrt{2}}{4}\int \frac{\sqrt{2}}{(\sqrt{2}x-1)^2+1}\,dx=\frac{\sqrt{2}}{4}\tan^{-1}(\sqrt{2}x-1)$

$\int \frac{2+\sqrt{2}x}{4(x^2+\sqrt{2}x+1)}\,dx=\int \frac{1+\sqrt{2}x}{4(x^2+\sqrt{2}x+1)}+\frac{1}{4(x^2+\sqrt{2}x+1)}\,dx$

$\int \frac{1+\sqrt{2}x}{4(x^2+\sqrt{2}x+1)}\,dx=\frac{\sqrt{2}}{8}\int \frac{2x+\sqrt{2}}{(x^2+\sqrt{2}x+1)}\,dx=\frac{\sqrt{2}}{8}\ln(x^2+\sqrt{2}x+1)$

$\int \frac{1}{4(x^2+\sqrt{2}x+1)}\,dx=\frac{\sqrt{2}}{4}\int \frac{\sqrt{2}}{(\sqrt{2}x+1)^2+1}\,dx=\frac{\sqrt{2}}{4}\tan^{-1}(\sqrt{2}x+1)$

$\therefore \int \frac{1}{x^4+1}\,dx=\frac{\sqrt{2}}{8}(\ln(x^2+\sqrt{2}x+1)-\ln(x^2-\sqrt{2}x+1)+2(\tan^{-1}(\sqrt{2}x-1)+\tan^{-1}(\sqrt{2}x+1)))$

$W^5$

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d3stiny

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Re: Correct method of integration?

« Reply #1 on: November 10, 2012, 11:43:11 pm »

+12

+ c

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BubbleWrapMan

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Re: Correct method of integration?

« Reply #2 on: November 11, 2012, 12:17:29 am »

Yeah, you can make it easier to follow though

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Truck

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Re: Correct method of integration?

« Reply #3 on: November 11, 2012, 12:19:58 am »

Yeah so I can safely say that'll never come up on a VCAA Exam 1... so don't worry too much about learning to integrate things that complicated lol.

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madoscar65

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Re: Correct method of integration?

« Reply #4 on: November 14, 2012, 01:53:37 pm »

Use u = x^2 at the start and it will simplify things

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Will T

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