find-arctan-1-x-1-dx-

Question Number 67017 by mathmax by abdo last updated on 21/Aug/19

f i n d \int a r c t a n (1 + \sqrt{x + 1}) d x

Commented by mathmax by abdo last updated on 24/Aug/19

l e t I = \int a r c t a n (1 + \sqrt{x + 1}) d x c h a n g e m e n t \sqrt{x + 1} = t g i v e x + 1 = t^{2} \Rightarrow

I = \int a r c t a n (1 + t) (2 t) d t = 2 \int t a r c t a n (t + 1) d t b y p a r t s

I = 2 {\frac{t^{2}}{2} a r c t a n (t + 1) - \int \frac{t^{2}}{2} \times \frac{1}{1 + {(t + 1)}^{2}} d t}

= t^{2} a r c t a n (t + 1) - \int \frac{t^{2}}{1 + t^{2} + 2 t + 1} d t

\int \frac{t^{2}}{t^{2} + 2 t + 2} d t = \int \frac{t^{2} + 2 t + 2 - 2 t - 2}{t^{2} + 2 t + 2} d t = t - \int \frac{2 t + 2}{t^{2} + 2 t + 2} d t

= t - l n (t^{2} + 2 t + 2) \Rightarrow I = t^{2} a r c t a n (t + 1) - t + l n (t^{2} + 2 t + 2) + c

I = (x + 1) a r c t a n (1 + \sqrt{x + 1}) - \sqrt{x + 1} + l n (x + 3 + 2 \sqrt{x + 1}) + c