What is the integral of (3x)/(x^4+1) ?

The integral of (3x)/(x^4+1) is 3/2 arctan(x^2)+C

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integral of (3x)/(x^4+1)

\int \frac{3 x}{x ^{4} + 1} d x

\frac{3}{2} arctan (x^{2}) + C

Solution steps

\int \frac{3 x}{x ^{4} + 1} d x

Take the constant out:

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

= 3 \cdot \int \frac{x}{x ^{4} + 1} d x

Apply u-substitution

: \int \frac{1}{2 ( u ^{2} + 1 )} d u

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

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}{u ^{2} + 1} d u

Use the common integral:

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

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

Substitute back

u = x^{2}

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

Simplify

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

= \frac{3}{2} arctan (x^{2})

Add a constant to the solution

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

\frac{1}{3 + 2 x}

\frac{s}{( s + 2 ) ( s ^{2} + 4 )}

\frac{\partial}{\partial x} (2 arctan (\frac{x}{y}))

y^{^{''''}} + 20 y^{^{''}} + 64 y = 0

\frac{d}{d x} (x^{2} arctan (6 x))

What is the integral of (3x)/(x^4+1) ?
The integral of (3x)/(x^4+1) is 3/2 arctan(x^2)+C