What is the integral of θarctan^2(θ) ?

The integral of θarctan^2(θ) is 1/2 θ^2arctan^2(θ)-θarctan(θ)+1/2 arctan^2(θ)+ln|sqrt(θ^2+1)|+C

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integral of θarctan^2(θ)

\int θ arctan^{2} (θ) d θ

\frac{1}{2} θ^{2} arctan^{2} (θ) - θ arctan (θ) + \frac{1}{2} arctan^{2} (θ) + ln θ^{2} + 1 + C

Solution steps

\int θ arctan^{2} (θ) d θ

Apply Integration By Parts

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

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

= \frac{1}{2} θ^{2} arctan^{2} (θ) - (- arctan^{2} (θ) + θ arctan (θ) + \frac{1}{2} arctan^{2} (θ) - ln 1 + θ^{2})

Simplify

\frac{1}{2} θ^{2} arctan^{2} (θ) - (- arctan^{2} (θ) + θ arctan (θ) + \frac{1}{2} arctan^{2} (θ) - ln 1 + θ^{2}) : \frac{1}{2} θ^{2} arctan^{2} (θ) - θ arctan (θ) + \frac{1}{2} arctan^{2} (θ) + ln θ^{2} + 1

= \frac{1}{2} θ^{2} arctan^{2} (θ) - θ arctan (θ) + \frac{1}{2} arctan^{2} (θ) + ln θ^{2} + 1

Add a constant to the solution

= \frac{1}{2} θ^{2} arctan^{2} (θ) - θ arctan (θ) + \frac{1}{2} arctan^{2} (θ) + ln θ^{2} + 1 + C

What is the integral of θarctan^2(θ) ?
The integral of θarctan^2(θ) is 1/2 θ^2arctan^2(θ)-θarctan(θ)+1/2 arctan^2(θ)+ln|sqrt(θ^2+1)|+C