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

The integral of ((arctan(x))^6)/(x^2+1) is (arctan^7(x))/7+C

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

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

\frac{arctan ^{7} ( x )}{7} + C

Solution steps

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

Apply u-substitution

: \int u^{6} d u

= \int u^{6} d u

Apply the Power Rule

: \frac{u ^{7}}{7}

= \frac{u ^{7}}{7}

Substitute back

u = arctan (x)

= \frac{arctan ^{7} ( x )}{7}

Add a constant to the solution

= \frac{arctan ^{7} ( x )}{7} + C

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What is the integral of ((arctan(x))^6)/(x^2+1) ?
The integral of ((arctan(x))^6)/(x^2+1) is (arctan^7(x))/7+C