What is the integral of ((3x)/(1+x^4)) ?

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

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integral of ((3x)/(1+x^4))

\int (\frac{3 x}{1 + x ^{4}}) d x

\frac{3}{2} arctan (x^{2}) + C

Solution steps

\int \frac{3 x}{1 + x ^{4}} d x

Take the constant out:

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

= 3 \cdot \int \frac{x}{1 + x ^{4}} d x

Apply u-substitution

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

= 3 \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

= 3 \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)

= 3 \cdot \frac{1}{2} arctan (u)

Substitute back

u = x^{2}

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

Simplify

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

= \frac{3}{2} arctan (x^{2})

Add a constant to the solution

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

\int_{0}^{2} (e^{x} - e^{- x})^{2} d x

\int x - 3 d x

y^{^{''}} - 2 y^{^{'}} + y = e^{x} arctan (x)

\int x sin (14 x) d x

x \to - \frac{π}{2} - lim (sec (x))

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