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

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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