11) yeah, euler's substitution is lengthy.
take √(x2+x+1)=t+x
then this will reduce to∫t2-t+1(t-2)(2t-1)dt
this can be solved easily now.
\hspace{-16}\mathbf{(1)\;\; \int\frac{1}{x+\sqrt{x^2+x+1}}dx}$\\\\\\ $\mathbf{(2)\;\; \int\frac{\sqrt{\sqrt{x^4+1}-x^2}}{x^4+1}dx}$
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6 Answers
71I suppose the first one uses an Euler's Substitution. But it goes very lengthy.
112) i suppose take x2=tan y
then the integration becomes √(secy-tany)√tanydy
i think this can be solved. not sure though
712) We take,
x2=tan y
=> 2x dx = sec2y dy
=> dx = sec2y2 . √tany dy
Now,
I = ∫ √(secy-tany)tan2y+1 . sec2y2 . √tany dy
I = √(secy-tany)2 . √tany dy
I = 12 . ∫ √cosec y -1 dy
Put cosec y -1 = z2 => -csc y cot y = 2z dz => dy = - 2z dz(1+z2).√z4+2z2
Hence,
I = - ∫ z2 dz(z2+1) . z . √z2+2
I = - ∫ z dz(z2+1) . √z2+2
Put z2+2 = p2
I = - ∫ p . dp(p2-1) . p
I = ∫ dp(-p2+1)
Which is direct formula.
