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

\int 3^{2} d x

f (x) = e^{\frac{z}{( z - 2 )}}

\int \frac{1}{( 3 x + 1 ) ^{4}} d x

\frac{\partial}{\partial x} (4 sin (x^{3} y^{2}))

\frac{( 3 u ^{2} )}{( u ^{2} + u ) ^{3}}

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)