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

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

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

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

2 x - 2

Solution steps

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

Apply the Power Rule

: [2 t^{\frac{1}{2}}]_{1}^{x}

= [2 t^{\frac{1}{2}}]_{1}^{x}

Apply radical rule:

= [2 t]_{1}^{x}

Compute the boundaries:

2 x - 2

= 2 x - 2

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

y = \frac{x - x}{x ^{2}}

ln (1 + 3 x)

\frac{d y}{d x} + 2 x y = y + 6 x - 3

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