What is the integral from 1 to 2 of 1/(8(2+2x-x^2)) ?

The integral from 1 to 2 of 1/(8(2+2x-x^2)) is (ln(1/(sqrt(3))+1)-ln(-1/(sqrt(3))+1))/(16sqrt(3))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 2 of 1/(8(2+2x-x^2))

\int_{1}^{2} \frac{1}{8 ( 2 + 2 x - x ^{2} )} d x

\frac{ln ( \frac{1}{3} + 1 ) - ln ( - \frac{1}{3} + 1 )}{16 3}

Decimal Notation

0.04752 \dots

Solution steps

\int_{1}^{2} \frac{1}{8 ( 2 + 2 x - x ^{2} )} d x

Take the constant out:

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

= \frac{1}{8} \cdot \int_{1}^{2} \frac{1}{2 + 2 x - x ^{2}} d x

Complete the square

2 + 2 x - x^{2} : - (x - 1)^{2} + 3

= \frac{1}{8} \cdot \int_{1}^{2} \frac{1}{- ( x - 1 ) ^{2} + 3} d x

Apply u-substitution

: \int_{0}^{1} \frac{1}{- u ^{2} + 3} d u

= \frac{1}{8} \cdot \int_{0}^{1} \frac{1}{- u ^{2} + 3} d u

Apply Integral Substitution

= \frac{1}{8} \cdot \int_{0}^{\frac{1}{3}} \frac{1}{3 ( - v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{8} \cdot \frac{1}{3} \cdot \int_{0}^{\frac{1}{3}} \frac{1}{- v ^{2} + 1} d v

Use the common integral:

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

= \frac{1}{8} \cdot \frac{1}{3} [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{0}^{\frac{1}{3}}

Simplify

\frac{1}{8} \cdot \frac{1}{3} [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{0}^{\frac{1}{3}} : \frac{1}{8 3} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{0}^{\frac{1}{3}}

= \frac{1}{8 3} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{0}^{\frac{1}{3}}

Compute the boundaries:

\frac{ln ( \frac{1}{3} + 1 ) - ln ( - \frac{1}{3} + 1 )}{2}

= \frac{1}{8 3} \cdot \frac{ln ( \frac{1}{3} + 1 ) - ln ( - \frac{1}{3} + 1 )}{2}

Simplify

= \frac{ln ( \frac{1}{3} + 1 ) - ln ( - \frac{1}{3} + 1 )}{16 3}

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

\frac{d}{d x} (A e^{x} + B e^{- x})

x y^{^{'}} = ((1 - x) y + x^{2} + x^{3}), y (3) = 2

x \to 3 - lim (\frac{x}{∣ x ∣})

\frac{\partial}{\partial x} (2 y^{7} + 7 y^{9} x^{6})

What is the integral from 1 to 2 of 1/(8(2+2x-x^2)) ?
The integral from 1 to 2 of 1/(8(2+2x-x^2)) is (ln(1/(sqrt(3))+1)-ln(-1/(sqrt(3))+1))/(16sqrt(3))