Can you help me.
integral from 4 to 1
x=( x^2-x+1)/(sqrt(x))
integral from pie to pie/4
x=secxtanx
integral from 4 to x
u=(2+(sqrt(u))^8
integral
x=sqrt(1+x^2x^5)
integral
x=tan^6xsec^4x
This one is easy. Just split up the fraction and use the properties of exponents to rewrite each term as a power of $\displaystyle x.$
$\displaystyle \int_{\pi/4}^\pi\sec x\tan x\,dx$
Well, what is the derivative of the secant? That should make the answer to this one obvious.
$\displaystyle \int_4^x(2+\sqrt u)^8\,du$
Substitute $\displaystyle v=2+\sqrt u\Rightarrow dv=\frac{du}{2\sqrt u},$ and then note that $\displaystyle \sqrt u=v-2.$
$\displaystyle \int\sqrt{1+x^2x^5}\,dx$
This has no elementary antiderivative. Did you write it correctly?
Rewrite, using the identity $\displaystyle \tan^2x+1=\sec^2x\colon$$\displaystyle \int\tan^6x\sec^4x\,dx$
$\displaystyle \int\tan^6x\sec^4x\,dx$
$\displaystyle =\int\tan^6x\sec^2x\sec^2x\,dx$
$\displaystyle =\int\tan^6x(\tan^2x+1)\sec^2x\,dx$
$\displaystyle =\int\tan^8x\sec^2x\,dx+\int\tan^6x\sec^2x\,dx.$
You should be able to take it from here.