What is the integral of (3x^3)/(4+x^8) ?

The integral of (3x^3)/(4+x^8) is 3/8 arctan((x^4)/2)+C

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integral of (3x^3)/(4+x^8)

\int \frac{3 x ^{3}}{4 + x ^{8}} d x

\frac{3}{8} arctan (\frac{x ^{4}}{2}) + C

Solution steps

\int \frac{3 x ^{3}}{4 + x ^{8}} 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 ^{3}}{4 + x ^{8}} d x

Apply u-substitution

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

= 3 \cdot \int \frac{1}{4 ( 4 + 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}{4} \cdot \int \frac{1}{4 + u ^{2}} d u

Apply Integral Substitution

= 3 \cdot \frac{1}{4} \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

= 3 \cdot \frac{1}{4} \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)

= 3 \cdot \frac{1}{4} \cdot \frac{1}{2} arctan (v)

Substitute back

= 3 \cdot \frac{1}{4} \cdot \frac{1}{2} arctan (\frac{x ^{4}}{2})

Simplify

3 \cdot \frac{1}{4} \cdot \frac{1}{2} arctan (\frac{x ^{4}}{2}) : \frac{3}{8} arctan (\frac{x ^{4}}{2})

= \frac{3}{8} arctan (\frac{x ^{4}}{2})

Add a constant to the solution

= \frac{3}{8} arctan (\frac{x ^{4}}{2}) + C

\frac{d x}{d y} = 3 x^{2} y^{3}

\int 3 x^{3} - 4 x^{2} + \frac{7}{x} - \frac{1}{x ^{2}} + 2 e^{x} d x

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{3 x} (x + 1)

f (x) = 9 x - 8 x, (1, 1)

\int \frac{e ^{3 x}}{e ^{3 x} - e ^{- 3 x}} d x

What is the integral of (3x^3)/(4+x^8) ?
The integral of (3x^3)/(4+x^8) is 3/8 arctan((x^4)/2)+C