What is the integral from 2 to 6 of 1/(sqrt(2x-3)) ?

The integral from 2 to 6 of 1/(sqrt(2x-3)) is 2

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integral from 2 to 6 of 1/(sqrt(2x-3))

\int_{2}^{6} \frac{1}{2 x - 3} d x

2

Solution steps

\int_{2}^{6} \frac{1}{2 x - 3} d x

Apply u-substitution

: \int_{1}^{3} 1 d u

= \int_{1}^{3} 1 d u

Integral of a constant:

\int a d x = a x

= [1 \cdot u]_{1}^{3}

Simplify

= [u]_{1}^{3}

Compute the boundaries:

2

= 2

x \to 0 + lim (\frac{3 ^{x} - 1}{x ^{2}})

\int \frac{x ^{3}}{a} d x

x 2 x^{2} + 7

\frac{\partial}{\partial x} (∣ x ∣ y + 2 x)

\frac{d}{d x} (3 e^{- 2 x})

What is the integral from 2 to 6 of 1/(sqrt(2x-3)) ?
The integral from 2 to 6 of 1/(sqrt(2x-3)) is 2