What is the integral from 2 to 6 of (1/(sqrt(x-2))) ?

The integral from 2 to 6 of (1/(sqrt(x-2))) is 4

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integral from 2 to 6 of (1/(sqrt(x-2)))

\int_{2}^{6} (\frac{1}{x - 2}) d x

4

Solution steps

\int_{2}^{6} \frac{1}{x - 2} d x

Apply u-substitution

: \int_{0}^{2} 2 d u

= \int_{0}^{2} 2 d u

Integral of a constant:

\int a d x = a x

= [2 u]_{0}^{2}

Compute the boundaries:

4

= 4

x \to 1 lim (ln ∣ ln (x) ∣)

\frac{d y}{d t} = 8 y

\frac{d y}{d x} = x^{3} + 2 x + π

\int_{- 1}^{2} (- x^{2} + x + 2) d x

\frac{d}{d x} (tan (x^{3}))

What is the integral from 2 to 6 of (1/(sqrt(x-2))) ?
The integral from 2 to 6 of (1/(sqrt(x-2))) is 4