What is the integral of (x^3)/(1+x^8) ?

The integral of (x^3)/(1+x^8) is 1/4 arctan(x^4)+C

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integral of (x^3)/(1+x^8)

\int \frac{x ^{3}}{1 + x ^{8}} d x

\frac{1}{4} arctan (x^{4}) + C

Solution steps

\int \frac{x ^{3}}{1 + x ^{8}} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{4} arctan (u)

Substitute back

u = x^{4}

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

Add a constant to the solution

= \frac{1}{4} arctan (x^{4}) + C

\frac{d}{d x} (\frac{1}{4} e^{- \frac{x}{2}})

x \to π lim (sin (x - π))

x \to 0 + lim (- \frac{3}{x})

y^{^{'}} = \frac{4 x ^{2} + y ^{2}}{x y}

\frac{d}{d x} (5 x^{3})

What is the integral of (x^3)/(1+x^8) ?
The integral of (x^3)/(1+x^8) is 1/4 arctan(x^4)+C