What is the integral from 0 to 2 of 1+5/(4+x^2) ?

The integral from 0 to 2 of 1+5/(4+x^2) is 2(1+(5pi)/(16))

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integral from 0 to 2 of 1+5/(4+x^2)

\int_{0}^{2} 1 + \frac{5}{4 + x ^{2}} d x

2 (1 + \frac{5 π}{16})

Decimal Notation

3.96349 \dots

Solution steps

\int_{0}^{2} 1 + \frac{5}{4 + x ^{2}} d x

Apply Integral Substitution

= \int_{0}^{1} 2 (1 + \frac{5}{4 + 4 u ^{2}}) 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} 1 + \frac{5}{4 + 4 u ^{2}} 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} 1 d u + \int_{0}^{1} \frac{5}{4 + 4 u ^{2}} d u)

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

\int_{0}^{1} \frac{5}{4 + 4 u ^{2}} d u = \frac{5 π}{16}

= 2 (1 + \frac{5 π}{16})

x \to - 8 + lim (\frac{x + 7}{x + 8})

x \to 0 + lim (x^{x} - 1)

\frac{d}{d x} (- 2 e^{- x} x)

sin (4 x)

\frac{\partial}{\partial x} (\frac{1}{3 x})

What is the integral from 0 to 2 of 1+5/(4+x^2) ?
The integral from 0 to 2 of 1+5/(4+x^2) is 2(1+(5pi)/(16))