What is the integral from 0 to 1 of 1/(4-x^2) ?

The integral from 0 to 1 of 1/(4-x^2) is (ln(3))/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 1 of 1/(4-x^2)

\int_{0}^{1} \frac{1}{4 - x ^{2}} d x

\frac{ln ( 3 )}{4}

Decimal

0.27465 \dots

Solution steps

\int_{0}^{1} \frac{1}{4 - x ^{2}} d x

Apply Integral Substitution

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

Simplify

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

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

Compute the boundaries:

\frac{ln ( 2 ) + ln ( \frac{3}{2} )}{2}

= \frac{1}{2} \cdot \frac{ln ( 2 ) + ln ( \frac{3}{2} )}{2}

Simplify

= \frac{ln ( 3 )}{4}

What is the integral from 0 to 1 of 1/(4-x^2) ?
The integral from 0 to 1 of 1/(4-x^2) is (ln(3))/4