Question

no idea

Answer 1

\[\int\limits_{0}^{1}\frac{ x^{21} }{ \sqrt{x^2-x+1} }\]

Answer 2

what i did
x^3+1=(x+1)(x^2-x+1)
x^2-x+1=(x^3-1)/(x+1)

Answer 3

anyone ??

Answer 4

\[\int\limits_{0}^{1}\frac{ x^{21} }{ \sqrt{\frac{ x^3+1 }{ x+1}} }\]

Answer 5

Try completing the square, then substituting.

Answer 6

i tried and reach nearly to last but in end mixed upppppp :(

Answer 7

x^21=(x^3)^7

Answer 8

also used this

Answer 9

\[x^2-x+1=x^2-x+\frac{1}{4}-\frac{1}{4}=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\]
Set \(u=x-\dfrac{1}{2}\), then \(x^{21}=\left(u+\dfrac{1}{2}\right)^{21}\).
\[\int\frac{\left(u+\dfrac{1}{2}\right)^{21}}{\sqrt{u^2-\dfrac{1}{4}}}~du\]
Trig sub would be next suggestion, then you'd either integrate by parts a bunch of times or use a power reduction formula.

Maybe there's a more efficient way using your work...

Answer 10

well i tried 10 to 12 times ..but in end no solution i m getting ....u plzzz try and if u get it thn give me a hint that will be enough