What is the integral of (4x^3)/(25+x^8) ?

The integral of (4x^3)/(25+x^8) is 1/5 arctan((x^4)/5)+C

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integral of (4x^3)/(25+x^8)

\int \frac{4 x ^{3}}{25 + x ^{8}} d x

\frac{1}{5} arctan (\frac{x ^{4}}{5}) + C

Solution steps

\int \frac{4 x ^{3}}{25 + 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}}{25 + x ^{8}} d x

Apply u-substitution

: \int \frac{1}{4 ( 25 + u ^{2} )} d u

= 4 \cdot \int \frac{1}{4 ( 25 + 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}{25 + u ^{2}} d u

Apply Integral Substitution

= 4 \cdot \frac{1}{4} \cdot \int \frac{1}{5 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 4 \cdot \frac{1}{4} \cdot \frac{1}{5} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 4 \cdot \frac{1}{4} \cdot \frac{1}{5} arctan (v)

Substitute back

= 4 \cdot \frac{1}{4} \cdot \frac{1}{5} arctan (\frac{x ^{4}}{5})

Simplify

4 \cdot \frac{1}{4} \cdot \frac{1}{5} arctan (\frac{x ^{4}}{5}) : \frac{1}{5} arctan (\frac{x ^{4}}{5})

= \frac{1}{5} arctan (\frac{x ^{4}}{5})

Add a constant to the solution

= \frac{1}{5} arctan (\frac{x ^{4}}{5}) + C

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What is the integral of (4x^3)/(25+x^8) ?
The integral of (4x^3)/(25+x^8) is 1/5 arctan((x^4)/5)+C