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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

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

y^{^{'}} + y^{3} = 0, y (0) = 3

\frac{\partial}{\partial x} (x^{3} y^{4} z^{2})

y = - \frac{2 x}{x ^{2} + 1}, (1, - 1)

f (x) = x^{3} - 3 x^{2} + 4 x - 7

\frac{d}{d v} (\frac{u}{u + v})

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