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

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

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

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

3 arctan (x^{2}) + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{1 + u ^{2}} 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

\frac{d x}{d t} = 349.344897 (1 - e^{- 0.0916 t})

\int (3 e^{- 0.4 x}) d x

\int 4 sin (x) - 1 + 8 x^{- 5} d x

f (- 3) = 20 x^{4}

\int 3 e^{4 x + e^{4 x}} d x

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