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

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

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integral of (4x^3)/(1+x^8)

\int \frac{4 x ^{3}}{1 + x ^{8}} d x

arctan (x^{4}) + C

Solution steps

\int \frac{4 x ^{3}}{1 + x ^{8}} d x

Take the constant out:

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

= 4 \cdot \int \frac{x ^{3}}{1 + x ^{8}} d x

Apply u-substitution

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

= 4 \cdot \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

= 4 \cdot \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)

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

Substitute back

u = x^{4}

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

Simplify

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

= arctan (x^{4})

Add a constant to the solution

= arctan (x^{4}) + C

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