What is the integral of sqrt(1/(4+x^2)) ?

The integral of sqrt(1/(4+x^2)) is ln(1/2 |sqrt(x^2+4)+x|)+C

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integral of sqrt(1/(4+x^2))

\int \frac{1}{4 + x ^{2}} d x

ln (\frac{1}{2} x^{2} + 4 + x) + C

Solution steps

\int \frac{1}{4 + x ^{2}} d x

Apply Integral Substitution

= \int 2 \frac{1}{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 \frac{1}{4 + 4 u ^{2}} d u

\frac{1}{4 + 4 u ^{2}} = \frac{1}{4} \frac{1}{1 + u ^{2}}

= 2 \cdot \int \frac{1}{4} \frac{1}{1 + u ^{2}} d u

Take the constant out:

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

= 2 \frac{1}{4} \cdot \int \frac{1}{1 + u ^{2}} d u

\frac{1}{4} = \frac{1}{2}

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{1 + u ^{2}} d u

Simplify

\frac{1}{1 + u ^{2}} : \frac{1}{1 + u ^{2}}

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{1 + u ^{2}} d u

Simplify

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{1 + u ^{2}} d u

Use the common integral:

\int \frac{1}{1 + u ^{2}} d u = ln u^{2} + 1 + u

= 2 \cdot \frac{1}{2} ln u^{2} + 1 + u

Substitute back

u = \frac{x}{2}

= 2 \cdot \frac{1}{2} ln (\frac{x}{2})^{2} + 1 + \frac{x}{2}

Simplify

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

= ln (\frac{1}{2} x^{2} + 4 + x)

Add a constant to the solution

= ln (\frac{1}{2} x^{2} + 4 + x) + C

\int x^{3} x + 10 d x

x \to 0 lim (tan (x^{2}))

\frac{d y}{d x} = \frac{2}{16 - x ^{2}}

\int (e^{1 - x}) d x

\frac{d}{d x} (\frac{1}{10} x - \frac{1}{8})

What is the integral of sqrt(1/(4+x^2)) ?
The integral of sqrt(1/(4+x^2)) is ln(1/2 |sqrt(x^2+4)+x|)+C