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

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

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

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

\frac{1}{2 3} arctan (3 x^{2}) + C

Solution steps

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

Apply u-substitution

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

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

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

Apply Integral Substitution

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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

f (x) = x - 6

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

\int_{- 2}^{0} (5 x^{2} + 5 x) d x

x \to \infty lim (5 x^{3})

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