What is the integral of (x^3)/(sqrt(9+x^8)) ?

The integral of (x^3)/(sqrt(9+x^8)) is 1/4 ln((|x^4+sqrt(9+x^8)|)/3)+C

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integral of (x^3)/(sqrt(9+x^8))

\int \frac{x ^{3}}{9 + x ^{8}} d x

\frac{1}{4} ln (\frac{x ^{4} + 9 + x ^{8}}{3}) + C

Solution steps

\int \frac{x ^{3}}{9 + x ^{8}} d x

Apply u-substitution

: \int \frac{1}{4 9 + u ^{2}} d u

= \int \frac{1}{4 9 + 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}{9 + 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}{3} x^{4})) + sec (arctan (\frac{1}{3} x^{4}))

Simplify

\frac{1}{4} ln tan (arctan (\frac{1}{3} x^{4})) + sec (arctan (\frac{1}{3} x^{4})) : \frac{1}{4} ln (\frac{x ^{4} + 9 + x ^{8}}{3})

= \frac{1}{4} ln (\frac{x ^{4} + 9 + x ^{8}}{3})

Add a constant to the solution

= \frac{1}{4} ln (\frac{x ^{4} + 9 + x ^{8}}{3}) + C

k = 1 \sum \infty 0. 5^{k} t

f (x) = - \frac{2}{x ^{2}}

20 + x - x^{2}, x^{2} - 5 x, [1, 3]

e^{5 x}, at x = \frac{1}{5} ln (3)

\int (x + 1) ln (x + 1) d x

What is the integral of (x^3)/(sqrt(9+x^8)) ?
The integral of (x^3)/(sqrt(9+x^8)) is 1/4 ln((|x^4+sqrt(9+x^8)|)/3)+C