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

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

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

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

\frac{1}{4} ln (\frac{x ^{4} + 25 + x ^{8}}{5}) + C

Solution steps

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

Apply u-substitution

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

= \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

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

Apply Trigonometric Substitution

: \int sec (v) d v

= \frac{1}{4} \cdot \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= \frac{1}{4} ln ∣ tan (v) + sec (v) ∣

Substitute back

= \frac{1}{4} ln tan (arctan (\frac{1}{5} x^{4})) + sec (arctan (\frac{1}{5} x^{4}))

Simplify

\frac{1}{4} ln tan (arctan (\frac{1}{5} x^{4})) + sec (arctan (\frac{1}{5} x^{4})) : \frac{1}{4} ln (\frac{x ^{4} + 25 + x ^{8}}{5})

= \frac{1}{4} ln (\frac{x ^{4} + 25 + x ^{8}}{5})

Add a constant to the solution

= \frac{1}{4} ln (\frac{x ^{4} + 25 + x ^{8}}{5}) + C

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

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