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

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

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 (2x^2)/(x^2+1)

\int_{0}^{1} \frac{2 x ^{2}}{x ^{2} + 1} d x

2 (- \frac{π}{4} + 1)

Decimal

0.42920 \dots

Solution steps

\int_{0}^{1} \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_{0}^{1} \frac{x ^{2}}{x ^{2} + 1} d x

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

= 2 \cdot \int_{0}^{1} - \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_{0}^{1} \frac{1}{x ^{2} + 1} d x + \int_{0}^{1} 1 d x)

\int_{0}^{1} \frac{1}{x ^{2} + 1} d x = \frac{π}{4}

\int_{0}^{1} 1 d x = 1

= 2 (- \frac{π}{4} + 1)

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