What is the integral of (6x)/(x^4+1) ?

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

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integral of (6x)/(x^4+1)

\int \frac{6 x}{x ^{4} + 1} d x

3 arctan (x^{2}) + C

Solution steps

\int \frac{6 x}{x ^{4} + 1} d x

Take the constant out:

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

= 6 \cdot \int \frac{x}{x ^{4} + 1} d x

Apply u-substitution

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

= 6 \cdot \int \frac{1}{2 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 6 \cdot \frac{1}{2} arctan (u)

Substitute back

u = x^{2}

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

Simplify

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

= 3 arctan (x^{2})

Add a constant to the solution

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

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