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

= 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

d er i v a t i v e f (x) = x^{3} - 3 x^{2} + 4 x - 7

\int cos (x) + 2 d x

\frac{d}{d x} (x (7 x + 9)^{\frac{1}{3}})

25 y^{^{''}} + 2 y = 0

t an g e n t f (x) = - x^{3}, (2, - 8)

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