1-tan-1-x-2-dx-2-tan-1-x-dx-

Question Number 117342 by bemath last updated on 11/Oct/20

(1) \int {(\tan^{- 1} (x))}^{2} dx = ?

(2) \int \tan^{- 1} (\sqrt{x}) dx = ?

Answered by mathmax by abdo last updated on 11/Oct/20

I = \int \arctan^{2} x dx by psrts I = x \arctan^{2} x - \int x \frac{(2 arctanx)}{1 + x^{2}} dx

= x \arctan^{2} x - \int \frac{2 x}{1 + x^{2}} \arctan (x) dx also by parts

\int \frac{2 x}{1 + x^{2}} \arctan (x) dx = \ln (1 + x^{2}) \arctan (x) - \int \frac{\ln (1 + x^{2})}{1 + x^{2}} dx

\int \frac{\ln (1 + x^{2})}{1 + x^{2}} dx =_{x = \tan θ} \int \frac{\ln (1 + \tan^{2} θ)}{1 + \tan^{2} θ} (1 + \tan^{2} θ) d θ

= \int \ln (\frac{1}{\cos^{2} θ}) d θ = - 2 \int \ln (\cos θ) d θ \Rightarrow

I = x \arctan^{2} x - \ln (1 + x^{2}) arctanx - 2 \int \ln (\cos θ) d θ \dots

Commented by mathmax by abdo last updated on 11/Oct/20

thank you but what mean ∣ \dots ∣ because at C

∣ {re}^{i θ} ∣ = r and ∣ x + iy ∣ = \sqrt{x^{2} + y^{2}} \dots!

Commented by bemath last updated on 11/Oct/20

the last term

\int \ln (\cos θ) d θ = ?

Commented by Lordose last updated on 11/Oct/20

check the pic i uploaded .

it can also be expressed using clausen

function of second order ({Cl}_{2})

Answered by bemath last updated on 11/Oct/20

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