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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 2 to 3 of (2x^2)/(x^2-1)

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

- ln (2) + ln (3) + 2

Decimal Notation

2.40546 \dots

Solution steps

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

Take the constant out:

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

= 2 \cdot \int_{2}^{3} \frac{x ^{2}}{x ^{2} - 1} d x

Factor

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

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

Take the constant out:

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

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

\frac{x ^{2}}{- x ^{2} + 1} = \frac{1}{- x ^{2} + 1} - 1

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int_{2}^{3} 1 d x = 1

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

Simplify

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

= - ln (2) + ln (3) + 2

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

\int_{- 1}^{1} \frac{8 x}{x ^{2} + 4} d x

\int_{1}^{13} (6 x^{2} + 3 x - 2) d x

\int_{0}^{\frac{π}{4}} tan (x^{2}) d x

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