What is the integral of (x^2)/(1+x^6) ?

The integral of (x^2)/(1+x^6) is 1/3 arctan(x^3)+C

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integral of (x^2)/(1+x^6)

\int \frac{x ^{2}}{1 + x ^{6}} d x

\frac{1}{3} arctan (x^{3}) + C

Solution steps

\int \frac{x ^{2}}{1 + x ^{6}} d x

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{1 + u ^{2}} d u = arctan (u)

= \frac{1}{3} arctan (u)

Substitute back

u = x^{3}

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

Add a constant to the solution

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

\int \frac{x ^{2}}{a} e^{- \frac{x}{a}} d x

\frac{\partial}{\partial x} (e^{(t - x)})

\frac{d}{d x} (4 x^{2} - 5)

y = 3 x^{3} - 10 x - \frac{4}{x ^{3}}

\frac{\partial}{\partial x} (\frac{3 x y}{x ^{4} + y ^{4}})

What is the integral of (x^2)/(1+x^6) ?
The integral of (x^2)/(1+x^6) is 1/3 arctan(x^3)+C