Question

. sq. rt of ( 2x - 1) = x

Answer 1

Make the substitution u = sqrt(1 + 2x).

Then du = (1 / sqrt(1 + 2x)) dx.

Solve the first equation for x:

u = sqrt(1 + 2x)
u^2 = 1 + 2x
u^2 - 1 = 2x
(u^2 - 1) / 2 = x.

Now you can plug in for x and dx in your integral:

∫ (x / sqrt(1 + 2x)) dx = ∫ [(u^2 - 1) / 2] du.

This can be integrated with the power rule:

∫ [(u^2 - 1) / 2] du = (1/2) ∫ (u^2 - 1) du = (1/2)(u^3 / 3 - u) + C = u^3 / 6 - u / 2 + C.

Now substitute back in for u:

u^3 / 6 - u / 2 + C = (sqrt(1 + 2x))^3 / 6 - (sqrt(1 + 2x)) / 2 + C.
= [(1 + 2x)^(3/2)] / 6 - (sqrt(1 + 2x)) / 2 + C.

Answer 2

i have done integration !!oops!!

Answer 3

so thats the wrong way ?

Answer 4

yep!

Answer 5

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Answer 6

\[\sqrt{2x-1}=x\]Square both sides getting\[2x-1=x ^{2}\] Transpose by subtracting (2x-1) from both sides.\[x ^{2}-2x+1=0\]

Answer 7

Factor getting (x-1)(x-1)=0
x=1 (double root)

Answer 8

check by substituting in original.\[\sqrt{2(1)-1}=1\]\[\sqrt{1}=1\]