What is the integral from 0 to a of 1/(a^2-x^2) ?

The integral from 0 to a of 1/(a^2-x^2) is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to a of 1/(a^2-x^2)

\int_{0}^{a} \frac{1}{a ^{2} - x ^{2}} d x

diverges

Solution steps

\int_{0}^{a} \frac{1}{a ^{2} - x ^{2}} d x

Factor

a^{2} - x^{2} : - (- a^{2} + x^{2})

= \int_{0}^{a} \frac{1}{- ( - a ^{2} + x ^{2} )} d x

Take the constant out:

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

= - \int_{0}^{a} \frac{1}{- a ^{2} + x ^{2}} d x

Apply Integral Substitution

= - \int_{0}^{1} \frac{1}{a ( u ^{2} - 1 )} d u

Take the constant out:

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

= - \frac{1}{a} \cdot \int_{0}^{1} \frac{1}{u ^{2} - 1} d u

Factor

u^{2} - 1 : - (- u^{2} + 1)

= - \frac{1}{a} \cdot \int_{0}^{1} \frac{1}{- ( - u ^{2} + 1 )} d u

Take the constant out:

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

= - \frac{1}{a} (- \int_{0}^{1} \frac{1}{- u ^{2} + 1} d u)

Use the common integral:

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= - \frac{1}{a} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{1})

Simplify

- \frac{1}{a} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{1}) : \frac{1}{a} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{0}^{1}

= \frac{1}{a} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{0}^{1}

Compute the boundaries:

diverges

= diverges

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

\int_{\frac{1}{2} π}^{π} 2 sin (x) d x

\int_{0}^{1 - x} (1 - x - y) d y

\int_{1}^{9} 10 x d x

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

What is the integral from 0 to a of 1/(a^2-x^2) ?
The integral from 0 to a of 1/(a^2-x^2) is diverges