What is the integral of (x^2)/(sqrt(49+x^6)) ?

The integral of (x^2)/(sqrt(49+x^6)) is 1/3 ln((|x^3+sqrt(49+x^6)|)/7)+C

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integral of (x^2)/(sqrt(49+x^6))

\int \frac{x ^{2}}{49 + x ^{6}} d x

\frac{1}{3} ln (\frac{x ^{3} + 49 + x ^{6}}{7}) + C

Solution steps

\int \frac{x ^{2}}{49 + x ^{6}} d x

Apply u-substitution

: \int \frac{1}{3 49 + u ^{2}} d u

= \int \frac{1}{3 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}{3} \cdot \int \frac{1}{49 + u ^{2}} d u

Apply Trigonometric Substitution

: \int sec (v) d v

= \frac{1}{3} \cdot \int sec (v) d v

Use the common integral:

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

= \frac{1}{3} ln ∣ tan (v) + sec (v) ∣

Substitute back

= \frac{1}{3} ln tan (arctan (\frac{1}{7} x^{3})) + sec (arctan (\frac{1}{7} x^{3}))

Simplify

\frac{1}{3} ln tan (arctan (\frac{1}{7} x^{3})) + sec (arctan (\frac{1}{7} x^{3})) : \frac{1}{3} ln (\frac{x ^{3} + 49 + x ^{6}}{7})

= \frac{1}{3} ln (\frac{x ^{3} + 49 + x ^{6}}{7})

Add a constant to the solution

= \frac{1}{3} ln (\frac{x ^{3} + 49 + x ^{6}}{7}) + C

y - x^{2}

n = 1 \sum \infty \frac{5}{n} - \frac{5}{n + 1}

\int_{0}^{\frac{π}{2}} 6 cos^{5} (x) d x

tan^{2} (3 x) + 1

ln (1 + \frac{1}{x ^{2} + 1})

What is the integral of (x^2)/(sqrt(49+x^6)) ?
The integral of (x^2)/(sqrt(49+x^6)) is 1/3 ln((|x^3+sqrt(49+x^6)|)/7)+C