What is the integral of 3/(u^2+2) ?

The integral of 3/(u^2+2) is 3/(sqrt(2))arctan(u/(sqrt(2)))+C

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integral of 3/(u^2+2)

\int \frac{3}{u ^{2} + 2} d u

\frac{3}{2} arctan (\frac{u}{2}) + C

Solution steps

\int \frac{3}{u ^{2} + 2} d u

Take the constant out:

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

= 3 \cdot \int \frac{1}{u ^{2} + 2} d u

Apply Integral Substitution

= 3 \cdot \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

= 3 \cdot \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)

= 3 \cdot \frac{1}{2} arctan (v)

Substitute back

v = \frac{u}{2}

= 3 \cdot \frac{1}{2} arctan (\frac{u}{2})

Simplify

3 \cdot \frac{1}{2} arctan (\frac{u}{2}) : \frac{3}{2} arctan (\frac{u}{2})

= \frac{3}{2} arctan (\frac{u}{2})

Add a constant to the solution

= \frac{3}{2} arctan (\frac{u}{2}) + C

f (x) = (2 x - x^{2})^{3}

\frac{d}{d x} (sec^{4} (x) - tan^{4} (x))

y^{^{'}} = 3 - (2 y)

\frac{\partial}{\partial y} (ln (x^{2} + y^{3}))

\frac{x - 5}{x}

What is the integral of 3/(u^2+2) ?
The integral of 3/(u^2+2) is 3/(sqrt(2))arctan(u/(sqrt(2)))+C