What is area (2x)/(x^2+1), x/(13),[-5,5] ?

The answer to area (2x)/(x^2+1), x/(13),[-5,5] is -25/13+2ln(26)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

area (2x)/(x^2+1), x/(13),[-5,5]

area

\frac{2 x}{x ^{2} + 1}, \frac{x}{13}, [- 5, 5]

- \frac{25}{13} + 2 ln (26)

Decimal

4.59311 \dots

Solution steps

Apply the area formula:

\int_{- 5}^{5} \frac{2 x}{x ^{2} + 1} - \frac{x}{13} d x

Solve

\int_{- 5}^{5} \frac{2 x}{x ^{2} + 1} - \frac{x}{13} d x : - \frac{25}{13} + 2 ln (26)

The area is:

= - \frac{25}{13} + 2 ln (26)

3 sin (x), 4 cos (x)

x + 5 x y - 2 x^{3} = 2

y^{^{''}} = x sin (x) + x cos (x)

\int x (x^{2} + 3) d x

n = 0 \sum \infty (n + 1) 0. 5^{n}

What is area (2x)/(x^2+1), x/(13),[-5,5] ?
The answer to area (2x)/(x^2+1), x/(13),[-5,5] is -25/13+2ln(26)