What is the integral of (x^2)/(sqrt(x^6+25)) ?

The integral of (x^2)/(sqrt(x^6+25)) is 1/3 ln((|x^3+sqrt(25+x^6)|)/5)+C

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integral of (x^2)/(sqrt(x^6+25))

\int \frac{x ^{2}}{x ^{6} + 25} d x

\frac{1}{3} ln (\frac{x ^{3} + 25 + x ^{6}}{5}) + C

Solution steps

\int \frac{x ^{2}}{x ^{6} + 25} d x

Apply u-substitution

= \int \frac{1}{3 u ^{2} + 25} 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}{u ^{2} + 25} d u

Apply Trigonometric Substitution

= \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}{5} x^{3})) + sec (arctan (\frac{1}{5} x^{3}))

Simplify

\frac{1}{3} ln tan (arctan (\frac{1}{5} x^{3})) + sec (arctan (\frac{1}{5} x^{3})) : \frac{1}{3} ln (\frac{x ^{3} + 25 + x ^{6}}{5})

= \frac{1}{3} ln (\frac{x ^{3} + 25 + x ^{6}}{5})

Add a constant to the solution

= \frac{1}{3} ln (\frac{x ^{3} + 25 + x ^{6}}{5}) + C

x \to 0 lim (x^{2} cos (x^{5}))

\frac{d}{d x} ((x^{\frac{1}{2}}) \cdot (x^{2} + x - 4))

\frac{d}{d x} (x^{3} + b x^{2} + c x + d)

\frac{\partial}{\partial y} (sin (x y) + x sin (y))

n = 0 \sum \infty (\frac{5}{5})^{n}

What is the integral of (x^2)/(sqrt(x^6+25)) ?
The integral of (x^2)/(sqrt(x^6+25)) is 1/3 ln((|x^3+sqrt(25+x^6)|)/5)+C