What is the integral from a to b of 1/((x-a)^p) ?

The integral from a to b of 1/((x-a)^p) is ((b-a)^{-p+1})/(-p+1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from a to b of 1/((x-a)^p)

\int_{a}^{b} \frac{1}{( x - a ) ^{p}} d x

\frac{( b - a ) ^{- p + 1}}{- p + 1}

Solution steps

\int_{a}^{b} \frac{1}{( x - a ) ^{p}} d x

Apply u-substitution

: \int_{0}^{- a + b} \frac{1}{u ^{p}} d u

= \int_{0}^{- a + b} \frac{1}{u ^{p}} d u

Apply the Power Rule

: [\frac{u ^{- p + 1}}{- p + 1}]_{0}^{- a + b}

= [\frac{u ^{- p + 1}}{- p + 1}]_{0}^{- a + b}

Compute the boundaries:

\frac{( b - a ) ^{- p + 1}}{- p + 1}

= \frac{( b - a ) ^{- p + 1}}{- p + 1}

\int_{0}^{t^{4}} u^{5} d u

\int_{5}^{12} \frac{1}{x - 5} d x

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

\int_{2}^{9} x - 1 d x

\int_{0}^{1} 2 x (- x + 1)^{3} d x

What is the integral from a to b of 1/((x-a)^p) ?
The integral from a to b of 1/((x-a)^p) is ((b-a)^{-p+1})/(-p+1)