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

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

Popular >

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

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

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

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u ^{2} + 1} 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 y} (x y z^{2})

f (w) = \frac{( 2 w ^{4} - 5 w )}{9 w}

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

\int 2 t - 4 d t

f (t) = 2 e^{- t} cos (4 t) 1 (t)

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