What is the integral of 1/((x-1)^{3/2)} ?

The integral of 1/((x-1)^{3/2)} is -2/((x-1)^{1/2)}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/((x-1)^{3/2)}

\int \frac{1}{( x - 1 ) ^{\frac{3}{2}}} d x

- \frac{2}{( x - 1 ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( x - 1 ) ^{\frac{3}{2}}} d x

Apply u-substitution

: \int \frac{1}{u ^{\frac{3}{2}}} d u

= \int \frac{1}{u ^{\frac{3}{2}}} d u

Apply the Power Rule

: - \frac{2}{u ^{\frac{1}{2}}}

= - \frac{2}{u ^{\frac{1}{2}}}

Substitute back

u = x - 1

= - \frac{2}{( x - 1 ) ^{\frac{1}{2}}}

Add a constant to the solution

= - \frac{2}{( x - 1 ) ^{\frac{1}{2}}} + C

\int_{0}^{4} \frac{1}{( 6 - 2 x ) ^{\frac{1}{3}}} d x

(\frac{y ^{^{'}} - e ^{- t} + 5}{y}) = - 5, y (0) = - 2

\int \frac{5}{9 - ( x - 3 ) ^{2}} d x

y = \frac{2}{x - 4}

\int \frac{1}{k - x} d x

What is the integral of 1/((x-1)^{3/2)} ?
The integral of 1/((x-1)^{3/2)} is -2/((x-1)^{1/2)}+C