What is the integral of x^7arctan(x^4) ?

The integral of x^7arctan(x^4) is 1/4 (1/2 x^8arctan(x^4)-1/2 (-arctan(x^4)+x^4))+C

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integral of x^7arctan(x^4)

\int x^{7} arctan (x^{4}) d x

\frac{1}{4} (\frac{1}{2} x^{8} arctan (x^{4}) - \frac{1}{2} (- arctan (x^{4}) + x^{4})) + C

Solution steps

\int x^{7} arctan (x^{4}) d x

Apply u-substitution

= \int \frac{u arctan ( u )}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int u arctan (u) d u

Apply Integration By Parts

= \frac{1}{4} (\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)

= \frac{1}{4} (\frac{1}{2} u^{2} arctan (u) - \frac{1}{2} (- arctan (u) + u))

Substitute back

u = x^{4}

= \frac{1}{4} (\frac{1}{2} (x^{4})^{2} arctan (x^{4}) - \frac{1}{2} (- arctan (x^{4}) + x^{4}))

(x^{4})^{2} = x^{8}

= \frac{1}{4} (\frac{1}{2} x^{8} arctan (x^{4}) - \frac{1}{2} (- arctan (x^{4}) + x^{4}))

Add a constant to the solution

= \frac{1}{4} (\frac{1}{2} x^{8} arctan (x^{4}) - \frac{1}{2} (- arctan (x^{4}) + x^{4})) + C

What is the integral of x^7arctan(x^4) ?
The integral of x^7arctan(x^4) is 1/4 (1/2 x^8arctan(x^4)-1/2 (-arctan(x^4)+x^4))+C