What is the integral of (x^3)/(sqrt(49+x^8)) ?

The integral of (x^3)/(sqrt(49+x^8)) is 1/4 ln((|x^4+sqrt(49+x^8)|)/7)+C

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integral of (x^3)/(sqrt(49+x^8))

\int \frac{x ^{3}}{49 + x ^{8}} d x

\frac{1}{4} ln (\frac{x ^{4} + 49 + x ^{8}}{7}) + C

Solution steps

\int \frac{x ^{3}}{49 + x ^{8}} d x

Apply u-substitution

: \int \frac{1}{4 49 + u ^{2}} d u

= \int \frac{1}{4 49 + 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}{49 + 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}{7} x^{4})) + sec (arctan (\frac{1}{7} x^{4}))

Simplify

\frac{1}{4} ln tan (arctan (\frac{1}{7} x^{4})) + sec (arctan (\frac{1}{7} x^{4})) : \frac{1}{4} ln (\frac{x ^{4} + 49 + x ^{8}}{7})

= \frac{1}{4} ln (\frac{x ^{4} + 49 + x ^{8}}{7})

Add a constant to the solution

= \frac{1}{4} ln (\frac{x ^{4} + 49 + x ^{8}}{7}) + C

\frac{d ^{2}}{d x ^{2}} (15 e^{x^{2}})

9 - x^{2}

\frac{\partial}{\partial x} ((2 x + 6 y)^{11})

\frac{11 x}{( 2 + x ^{2} )}

x \to 2 lim (\frac{x + 1}{x - 1})

What is the integral of (x^3)/(sqrt(49+x^8)) ?
The integral of (x^3)/(sqrt(49+x^8)) is 1/4 ln((|x^4+sqrt(49+x^8)|)/7)+C