1-x-x-1-dx-

Question Number 168745 by MikeH last updated on 17/Apr/22

\int \frac{1}{x + \sqrt{x - 1}} d x = ? ?

Commented by safojontoshtemirov last updated on 17/Apr/22

\int \frac{d x}{x + \sqrt{x - 1}} = \sqrt{x - 1} = t; x = t^{2} + 1; d x = 2 t d t

\int \frac{2 t d t}{t^{2} + t + 1} = \int \frac{2 t + 1}{t^{2} + t + 1} d t - \int \frac{1}{t^{2} + t + 1} d t = I_{1} - I_{2}

I_{1} = \int \frac{d (t^{2} + t + 1)}{t^{2} + t + 1} d t = l n (t^{2} + t + 1) + C_{1} = l n (x - 1 + \sqrt{x - 1} + 1) + C_{1}

I_{2} = \int \frac{1}{t^{2} + t + 1} d t = \int \frac{1}{{(t + \frac{1}{2})}^{2} + \frac{3}{4}} d t = \frac{4}{3} \int \frac{1}{{(\frac{t + \frac{1}{2}}{\frac{\sqrt{3}}{2}})}^{2} + 1} d t =

\frac{4}{3} \int \frac{1}{{(\frac{2 t + 1}{\sqrt{3}})}^{2} + 1} d t = \frac{4}{3} \cdot \frac{\sqrt{3}}{2} \int \frac{d (\frac{2 t + 1}{\sqrt{3}})}{{(\frac{2 t + 1}{\sqrt{3}})}^{2} + 1} =

\frac{1}{\sqrt{3}} a r c t g (\frac{2 t + 1}{\sqrt{3}}) + C_{2} = \frac{2}{\sqrt{3}} a r c t g \frac{2 \sqrt{x - 1} + 1}{\sqrt{3}} + C_{2}

I_{1} - I_{2} = l n (x - 1 + \sqrt{x - 1} + 1) - \frac{2}{\sqrt{3}} a r c t g \frac{2 \sqrt{x - 1} + 1}{\sqrt{3}} + C

Answered by blackmamba last updated on 17/Apr/22

= \int \frac{x - \sqrt{x - 1}}{x^{2} - x + 1} d x

= \int \frac{\frac{1}{2} (2 x - 1) + \frac{1}{2}}{x^{2} - x + 1} d x - \int \frac{\sqrt{x - 1}}{x^{2} - x + 1} d x

= \frac{1}{2} \ln (x^{2} - x + 1) + \frac{1}{2} \int \frac{d x}{{(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} - \int \frac{\sqrt{x - 1}}{x^{2} - x + 1} d x