What is the integral of ((arctan(x))^4)/(1+x^2) ?

The integral of ((arctan(x))^4)/(1+x^2) is (arctan^5(x))/5+C

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integral of ((arctan(x))^4)/(1+x^2)

\int \frac{( arctan ( x ) ) ^{4}}{1 + x ^{2}} d x

\frac{arctan ^{5} ( x )}{5} + C

Solution steps

\int \frac{( arctan ( x ) ) ^{4}}{1 + x ^{2}} d x

Apply u-substitution

: \int u^{4} d u

= \int u^{4} d u

Apply the Power Rule

: \frac{u ^{5}}{5}

= \frac{u ^{5}}{5}

Substitute back

u = arctan (x)

= \frac{arctan ^{5} ( x )}{5}

Add a constant to the solution

= \frac{arctan ^{5} ( x )}{5} + C

\frac{\partial}{\partial x} (y ln (x^{3} + y^{4} + 2))

\frac{d}{d x} (e^{x z (x)})

f (x) = x^{2} + 169 x + 820, at x = 2

y = \frac{2 x}{7 - cot ( x )}

\int (x^{- \frac{1}{2}}) d x

What is the integral of ((arctan(x))^4)/(1+x^2) ?
The integral of ((arctan(x))^4)/(1+x^2) is (arctan^5(x))/5+C