please help me solve the improper integral of the integral from 0 to 2 of (2x+1)/square root of(x2-x

Question

anonymous · Answer

substitute u = x^2-x
du = 2x-1

anonymous · Answer

is the denominator x^2+x or x^2-x ?

anonymous · Answer

the sign of the denominator is a plus

anonymous · Answer

then do what I suggested.
u = x^2+x
du = 2x+1

anonymous · Answer

so it becomes the integral from 0 to2 of du/square root of u which gives2(x2-1)1/2 from c to2?

anonymous · Answer

if u = x^2+x,
then at x =0, u =0
and x =2, u = 6
so your integral becomes
$$\int\limits_{0}^{6}\frac{du}{\sqrt u}$$

anonymous · Answer

thanks pls, what of reduced formulas?

anonymous · Answer

huh?

anonymous · Answer

reduced formulas? what do you mean?

anonymous · Answer

fine am saying i have a problem in finding reduced formulas of trigonometric fxns in integration.

anonymous · Answer

then you should post it. vague questions will get you vague answers

anonymous · Answer

what is the reduced formula for costx power n  dx?

anonymous · Answer

use equation button below to type out your question
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