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

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

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 2x(1-1/(x^2))

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

3 - 2 ln (2)

Decimal Notation

1.61370 \dots

Solution steps

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

Take the constant out:

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

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

Multiply by the conjugate of

1 - \frac{1}{x ^{2}} : \frac{1 - \frac{1}{x ^{4}}}{1 + \frac{1}{x ^{2}}}

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

Simplify

x \frac{1 - \frac{1}{x ^{4}}}{1 + \frac{1}{x ^{2}}} : \frac{( x + 1 ) ( x - 1 )}{x}

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

Expand

\frac{( x + 1 ) ( x - 1 )}{x} : x - \frac{1}{x}

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

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

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

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

Simplify

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

= 3 - 2 ln (2)

f (x) = (3 t - 1)^{4} (2 t + 1)^{- 3}

\int \frac{1}{( 4 - x ) ^{\frac{3}{2}}} d x

n = 1 \sum \infty \frac{5 n}{4 n + 7}

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