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

The integral of (arctan^5(x))/(1+x^2) is (arctan^6(x))/6+C

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integral of (arctan^5(x))/(1+x^2)

\int \frac{arctan ^{5} ( x )}{1 + x ^{2}} d x

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

Solution steps

\int \frac{arctan ^{5} ( x )}{1 + x ^{2}} d x

Apply u-substitution

: \int u^{5} d u

= \int u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= \frac{u ^{6}}{6}

Substitute back

u = arctan (x)

= \frac{arctan ^{6} ( x )}{6}

Add a constant to the solution

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

\frac{d y}{d x}, 9 x = y^{2}

x \to 0 + lim ((1 + 4 x)^{\frac{3}{x}})

\int_{0}^{5} - x^{2} + 5 x d x

\frac{d y}{d x} = y + x

\int_{0}^{2} t^{3} d t

What is the integral of (arctan^5(x))/(1+x^2) ?
The integral of (arctan^5(x))/(1+x^2) is (arctan^6(x))/6+C