What is the integral from-a to a of 1/((s-x)^2) ?

The integral from-a to a of 1/((s-x)^2) is (2a)/((s-a)(a+s))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-a to a of 1/((s-x)^2)

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

\frac{2 a}{( s - a ) ( a + s )}

Solution steps

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

Apply u-substitution

: \int_{s + a}^{s - a} - \frac{1}{u ^{2}} d u

= \int_{s + a}^{s - a} - \frac{1}{u ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int_{s + a}^{s - a} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{s + a}^{s - a}

= - [- \frac{1}{u}]_{s + a}^{s - a}

Compute the boundaries:

- \frac{1}{s - a} + \frac{1}{s + a}

= - (- \frac{1}{s - a} + \frac{1}{s + a})

Simplify

= \frac{2 a}{( s - a ) ( a + s )}

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

\int \frac{2}{x ^{\frac{2}{3}}} d x

2 y \cdot \frac{d y}{d x} = x (y^{2} + 1)

\frac{d}{d x} (in (x^{2} - 5))

y^{^{'}} = 1 - e^{x - y + 8}

What is the integral from-a to a of 1/((s-x)^2) ?
The integral from-a to a of 1/((s-x)^2) is (2a)/((s-a)(a+s))