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

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

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

\int \frac{x ^{2} arctan ( x ^{3} )}{1 + x ^{6}} d x

\frac{1}{6} arctan^{2} (x^{3}) + C

Solution steps

\int \frac{x ^{2} arctan ( x ^{3} )}{1 + x ^{6}} d x

Apply u-substitution

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

Apply u-substitution

= \frac{1}{3} \cdot \int v d v

Apply the Power Rule

= \frac{1}{3} \cdot \frac{v ^{2}}{2}

Substitute back

= \frac{1}{3} \cdot \frac{arctan ^{2} ( x ^{3} )}{2}

Simplify

\frac{1}{3} \cdot \frac{arctan ^{2} ( x ^{3} )}{2} : \frac{1}{6} arctan^{2} (x^{3})

= \frac{1}{6} arctan^{2} (x^{3})

Add a constant to the solution

= \frac{1}{6} arctan^{2} (x^{3}) + C

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