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

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

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integral from 0 to 2 of (x^2)/(x^2+4)

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

2 (- \frac{π}{4} + 1)

Decimal Notation

0.42920 \dots

Solution steps

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

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

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

Apply Integral Substitution

= \int_{0}^{1} 2 (- \frac{1}{u ^{2} + 1} + 1) d u

Take the constant out:

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

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

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}{u ^{2} + 1} d u + \int_{0}^{1} 1 d u)

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

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

= 2 (- \frac{π}{4} + 1)

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

y d x = (y e^{y - 2 x}) d y

f (x) = 6 - 7 x

x \to \infty lim (\frac{5 x + 6 x ^{2}}{3 x - 8})

\frac{1}{2}

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