Question

how to integrate this

Answer 1

\[\int\limits \frac{ x^2+x+1 }{ \sqrt{x^2-x+1} } dx\]

Answer 2

for denominator you will go like
\[\int\limits_{}^{} 1/x^2-x+1/4-1/4+1\]
\[\int\limits_{}^{} 1/(x-1/2)^2 +3/4\]
\[\int\limits_{}^{}4/3 (1/x-1/2)^2/3/4 +1\]
\[4/3 \int\limits_{}^{} (1/x-1/2/\sqrt{3}/2)^2 +1\]
let \[x-1/2/\sqrt{3}/2= \tan \theta \]
derivate x wrt to theta and substitute

Answer 3

@aryandecoolest : sorry ... i did not understand ..

Answer 4

I am thinking somehting like below :
if we let \(u = \sqrt{x^2-x+1}\),

then \(du = \dfrac{2x-1}{2\sqrt{x^2-x+1}}dx\)

Answer 5

so i would like to have 2x-1 on numerator :

\[\begin{align}\\ \int\limits \frac{ x^2+x+1 }{ \sqrt{x^2-x+1} } dx &= \int\limits \frac{(2x-1)+ x^2-x+2 }{ \sqrt{x^2-x+1} } dx \\~\\ &= \int\limits \frac{(2x-1)dx}{ \sqrt{x^2-x+1} } + \int\limits \frac{ (x^2-x+1 )dx}{ \sqrt{x^2-x+1} } + \int\limits \frac{ dx }{ \sqrt{x^2-x+1} } \\~\\ \end{align} \]

Answer 6

see if that looks okay ^^

Answer 7

i too did till this point. but looking at the answer given in the book i felt i was not correct. probably the answer printed in incorrect. @ganeshie8 .. thanks a lot

Answer 8

you're on right track, don't wry about the answer yet...
your answer will come out same as textbook answer im sure, keep going :)