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

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

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

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

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

Solution steps

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

Take the constant out:

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

= 4 \cdot \int \frac{x}{x ^{4} + 4} d x

Apply u-substitution

: \int \frac{1}{2 ( u ^{2} + 4 )} d u

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

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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