What is the integral of x/(x^4+4x^2+8) ?

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

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integral of x/(x^4+4x^2+8)

\int \frac{x}{x ^{4} + 4 x ^{2} + 8} d x

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

Solution steps

\int \frac{x}{x ^{4} + 4 x ^{2} + 8} d x

Apply u-substitution

: \int \frac{1}{4 u ( 4 u + u + 8 )} d u

= \int \frac{1}{4 u ( 4 u + u + 8 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int \frac{1}{u ( 4 u + u + 8 )} d u

Apply u-substitution

: \int \frac{2}{4 v + v ^{2} + 8} d v

= \frac{1}{4} \cdot \int \frac{2}{4 v + v ^{2} + 8} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot 2 \cdot \int \frac{1}{4 v + v ^{2} + 8} d v

Complete the square

4 v + v^{2} + 8 : (v + 2)^{2} + 4

= \frac{1}{4} \cdot 2 \cdot \int \frac{1}{( v + 2 ) ^{2} + 4} d v

Apply u-substitution

: \int \frac{1}{w ^{2} + 4} d w

= \frac{1}{4} \cdot 2 \cdot \int \frac{1}{w ^{2} + 4} d w

Apply Integral Substitution

= \frac{1}{4} \cdot 2 \cdot \int \frac{1}{2 ( t ^{2} + 1 )} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot 2 \cdot \frac{1}{2} \cdot \int \frac{1}{t ^{2} + 1} d t

Use the common integral:

\int \frac{1}{t ^{2} + 1} d t = arctan (t)

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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