Question

Prove that

Answer 1

\[\frac{d}{dx}(\frac{1}{4 \sqrt{2}}\log| \frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}|+\frac{1}{2 \sqrt{2}} \tan^{-1}(\frac{\sqrt{2}x}{1-x^2}))=\frac{1}{1+x^4}\]

Answer 2

integrating RHS hmm

Answer 3

take x^2 common from denominator .
then let 1/x = t
See if this helps.

Answer 4

hey we have to differentiate LHS to prove it to RHS :| this question is from chapter 5 :|

Answer 5

and 1/1+x^4 to kal he integrate kiya tha

Answer 6

$$\frac{d}{dx}\left[\frac1{4\sqrt2}\log(x^2+\sqrt2x +1)\right]=\frac14\frac{\sqrt2x+1}{x^2+\sqrt2x+1}=\frac14\left(1-\frac{x^2}{x^2+\sqrt2x+1}\right)\\\frac{d}{dx}\left[\frac1{4\sqrt2}\log(x^2-\sqrt2x+1)\right]=\frac14\frac{\sqrt2x-1}{x^2-\sqrt2x+1}=\frac14\left(\frac{x^2}{x^2-\sqrt2x+1}-1\right)\\\frac{d}{dx}\left[\frac1{2\sqrt2}\arctan\frac{\sqrt2x}{1-x^2}\right]=\frac12\frac{(1-x^2)^2+2x^2(1-x^2)^2}{(1-x^2)^2+2x^2}=\frac12\frac{1+x^4-4x^4+2x^6}{1+x^4}$$etc.

Answer 7

$$\frac12\left(1-\frac{x^2}{x^2+\sqrt2x+1}-\frac{x^2}{x^2-\sqrt2x+1}+1\right)=1-\frac{\sqrt2x^3}{1+x^4}$$

Answer 8

wait that's not right:$$\frac12\left(1-\frac{x^2}{x^2+\sqrt2x+1}-\frac{x^2}{x^2-\sqrt2x+1}+1\right)=1-\frac{x^2+x^4}{1+x^4}$$

Answer 9

oops, I never finished working out the algebra. Try resimplifying the last derivative and sum them.