What is the integral of x/(x^4+16) ?

The integral of x/(x^4+16) is 1/8 arctan((x^2)/4)+C

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integral of x/(x^4+16)

\int \frac{x}{x ^{4} + 16} d x

\frac{1}{8} arctan (\frac{x ^{2}}{4}) + C

Solution steps

\int \frac{x}{x ^{4} + 16} d x

Apply u-substitution

: \int \frac{1}{2 ( u ^{2} + 16 )} d u

= \int \frac{1}{2 ( u ^{2} + 16 )} 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} + 16} d u

Apply Integral Substitution

= \frac{1}{2} \cdot \int \frac{1}{4 ( 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}{4} \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}{4} arctan (v)

Substitute back

= \frac{1}{2} \cdot \frac{1}{4} arctan (\frac{x ^{2}}{4})

Simplify

\frac{1}{2} \cdot \frac{1}{4} arctan (\frac{x ^{2}}{4}) : \frac{1}{8} arctan (\frac{x ^{2}}{4})

= \frac{1}{8} arctan (\frac{x ^{2}}{4})

Add a constant to the solution

= \frac{1}{8} arctan (\frac{x ^{2}}{4}) + C

2 y^{^{'''}} - 3 y^{^{''}} - 18 y^{^{'}} - 8 y = 0

y = 3 x^{2} - 12, y = 0

\frac{d}{d x} (\frac{3 + sin ( x )}{3 x + cos ( x )})

y = e^{x}, y = e^{- 3 x}, x = - 2, x = 1

\frac{d}{d x} (e^{3 (x - 1)})

What is the integral of x/(x^4+16) ?
The integral of x/(x^4+16) is 1/8 arctan((x^2)/4)+C