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

The integral of 1/(4+u^2) is 1/2 arctan(u/2)+C

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integral of 1/(4+u^2)

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

\frac{1}{2} arctan (\frac{u}{2}) + C

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{2 ( v ^{2} + 1 )} d v

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= \frac{1}{2} arctan (v)

Substitute back

v = \frac{u}{2}

= \frac{1}{2} arctan (\frac{u}{2})

Add a constant to the solution

= \frac{1}{2} arctan (\frac{u}{2}) + C

x y^{^{'}} + y = e^{x}, y (1) = 2

\frac{d}{d x} (cos^{2} (x) - sin^{2} (x))

\frac{d}{d x} (arctan (\frac{x + 1}{x - 1}))

\int \frac{1}{( x + 1 ) ^{3} ( x + 2 )} d x

f (x) = (2 x - 2)^{4} (x^{2} + x + 1)^{5}

What is the integral of 1/(4+u^2) ?
The integral of 1/(4+u^2) is 1/2 arctan(u/2)+C