What is the integral of x/(x^4+9) ?

The integral of x/(x^4+9) is 1/6 arctan((x^2)/3)+C

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integral of x/(x^4+9)

\int \frac{x}{x ^{4} + 9} d x

\frac{1}{6} arctan (\frac{x ^{2}}{3}) + C

Solution steps

\int \frac{x}{x ^{4} + 9} d x

Apply u-substitution

: \int \frac{1}{2 ( u ^{2} + 9 )} d u

= \int \frac{1}{2 ( u ^{2} + 9 )} 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} + 9} 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}{6} arctan (\frac{x ^{2}}{3})

= \frac{1}{6} arctan (\frac{x ^{2}}{3})

Add a constant to the solution

= \frac{1}{6} arctan (\frac{x ^{2}}{3}) + C

\int \frac{x + 4}{x ^{2} + 2 x + 5} d x

\int_{0}^{2} x d x

f (x) = - 3 x^{3} - 3 x^{2} - x + 3, at x = - 1

\int e^{3 x - 1} d x

x \to - 4 lim (x^{2} - 5 x + 12)

What is the integral of x/(x^4+9) ?
The integral of x/(x^4+9) is 1/6 arctan((x^2)/3)+C