What is the integral of 24x^5arctan(2x^3) ?

The integral of 24x^5arctan(2x^3) is 4x^6arctan(2x^3)-2x^3+arctan(2x^3)+C

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integral of 24x^5arctan(2x^3)

\int 24 x^{5} arctan (2 x^{3}) d x

4 x^{6} arctan (2 x^{3}) - 2 x^{3} + arctan (2 x^{3}) + C

Solution steps

\int 24 x^{5} arctan (2 x^{3}) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 24 \cdot \int x^{5} arctan (2 x^{3}) d x

Apply u-substitution

: \int \frac{u arctan ( u )}{12} d u

= 24 \cdot \int \frac{u arctan ( u )}{12} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 24 \cdot \frac{1}{12} \cdot \int u arctan (u) d u

Apply Integration By Parts

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

= 24 \cdot \frac{1}{12} (\frac{1}{2} u^{2} arctan (u) - \int \frac{u ^{2}}{2 ( u ^{2} + 1 )} d u)

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

= 24 \cdot \frac{1}{12} (\frac{1}{2} u^{2} arctan (u) - \frac{1}{2} (- arctan (u) + u))

Substitute back

u = 2 x^{3}

= 24 \cdot \frac{1}{12} (\frac{1}{2} (2 x^{3})^{2} arctan (2 x^{3}) - \frac{1}{2} (- arctan (2 x^{3}) + 2 x^{3}))

Simplify

24 \cdot \frac{1}{12} (\frac{1}{2} (2 x^{3})^{2} arctan (2 x^{3}) - \frac{1}{2} (- arctan (2 x^{3}) + 2 x^{3})) : 4 x^{6} arctan (2 x^{3}) - 2 x^{3} + arctan (2 x^{3})

= 4 x^{6} arctan (2 x^{3}) - 2 x^{3} + arctan (2 x^{3})

Add a constant to the solution

= 4 x^{6} arctan (2 x^{3}) - 2 x^{3} + arctan (2 x^{3}) + C

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What is the integral of 24x^5arctan(2x^3) ?
The integral of 24x^5arctan(2x^3) is 4x^6arctan(2x^3)-2x^3+arctan(2x^3)+C