39are they doubts?
\hspace{-16}\mathbf{(1)::\int\frac{1}{x+\sqrt{x^2+x+1}}dx}\\\\\\ \mathbf{(2)::\int\frac{x^{2001}}{(1+x^2)^{1002}}dx}\\\\\\ \mathbf{(3)::\int\frac{x^3-x}{x^6+4x^4+4x^2+1}dx}
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1\int \frac{x^{2001}dx}{(1+x^{2})^{1002}}
\int \frac{x^{2001}dx}{x^{2004}(1+\frac{1}{x^{2}})^{1002}}
\int \frac{x^{-3}dx}{(1+\frac{1}{x^{2}})^{1002}}
let\, \, 1+\frac{1}{x^{2}}=t\Rightarrow {-2}x^{-3} dx = dt
integral\, \, is\, \, \frac{-1}{2}\int \frac{dt}{t^{1002}}=\frac{t^{-1001}}{2002}
71For first, I'm stuck at \int \frac{\sqrt{x^2+x+1}}{x+1} \;dx
However, I worked with general substitutions? May you post the solutions please man111?
1Euler's SUBSTITUTION::
\sqrt{x^2+x+1}=tx+1\Rightarrow x^2+x+1= t^2x^2+2tx+1
x^2(1-t^2)=(2t-1)x\Rightarrow x=\frac{2t-1}{1-t^2}
differentiating both sides we get
dx =\frac{2(t^2-t+1)}{(1-t^2)^2} dt
x+\sqrt{x^2+x+1}=x(t+1)+1=\frac{(2t-1)(t+1)}{(1-t)(1+t)}+1 =\frac{t}{1-t}
the integral becomes ::
I=\int \frac{2(t^2-t+1)}{(1-t^2)^2(\frac{t}{1-t})} dt= \int \frac{2(t^2-t+1)}{(1-t)(1+t)^2t} dt
which can be done using partial fractions :::
1708\hspace{-16}\mathbf{\int\frac{\sqrt{x^2+x+1}}{x+1}dx=\int\frac{x(x+1)+1}{(x+1)\sqrt{x^2+x+1}}dx}$\\\\\\ $\mathbf{\int\frac{x}{\sqrt{x^2+x+1}}+\int\frac{1}{(x+1)\sqrt{x^2+x+1}}dx}$\\\\\\ Let $\mathbf{I_{1}=\int \frac{x}{\sqrt{x^2+x+1}}dx}$\\\\\\ and $\mathbf{I_{2}=\int \frac{1}{(x+1)\sqrt{x^2+x+1}}dx}$\\\\\\ Put $\mathbf{x+1=\frac{1}{t}\Leftrightarrow dx =-\frac{1}{t^2}dt}$
But I also think that euler substution is Nice......