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

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

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

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

2 - 2 t - 1

Solution steps

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

Apply u-substitution

: \int_{t - 1}^{1} 2 d u

= \int_{t - 1}^{1} 2 d u

Integral of a constant:

\int a d x = a x

= [2 u]_{t - 1}^{1}

Compute the boundaries:

2 - 2 t - 1

= 2 - 2 t - 1

\frac{d}{d x} (- (\frac{1}{x}))

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

cos (sin (x))

\frac{\partial}{\partial x} (e^{\frac{- 2 y}{x}})

\frac{d}{d x} (\frac{e ^{2 x}}{x ^{4}})

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