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

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

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integral of x/(x^4+3)

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

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

Solution steps

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

Apply u-substitution

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{12 s + 8}{( s - 1 ) ^{5}}

\frac{3}{x ^{3}} - \frac{1}{x ^{4}}

(26 \frac{2}{3}, 9), (5, 2.5)

x \to 0 lim (x^{2} cos (x))

f (x) = \frac{5}{x}, at x = 5

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