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

Question Number 144985 by imjagoll last updated on 01/Jul/21

\int \frac{(2 + \sqrt{x})}{{(x + 1 + \sqrt{x})}^{2}} dx = ?

Answered by mathmax by abdo last updated on 01/Jul/21

f (a) = \int \frac{2 + \sqrt{x}}{x + a + \sqrt{x}} with a > \frac{1}{4} \Rightarrow f^{^{'}} (a) = - \int \frac{2 + \sqrt{x}}{{(x + a + \sqrt{x})}^{2}} dx \Rightarrow

- f^{^{'}} (1) = \int \frac{2 + \sqrt{x}}{{(x + 1 + \sqrt{x})}^{2}} dx we have

f (a) =_{\sqrt{x} = t} \int \frac{2 + t}{t^{2} + a + t} (2 t) dt = \int \frac{2 t + t^{2}}{t^{2} + t + a} dt

= \int \frac{t^{2} + t + a + t - a}{t^{2} + t + a} dt = t + \int \frac{t - a}{t^{2} + t + a} dt

t^{2} + t + a = 0 \Rightarrow Δ = 1 - 4 a < 0 because a > \frac{1}{4} \Rightarrow

\int \frac{t - a}{t^{2} + t + a} dt = \frac{1}{2} \int \frac{2 t + 1 - 2 a - 1}{t^{2} + t + a} dt

= \frac{1}{2} \int \frac{2 t + 1}{t^{2} + t + a} dt - \frac{2 a + 1}{2} \int \frac{dt}{t^{2} + t + a}

= \frac{1}{2} \log (t^{2} + t + a) - \frac{2 a + 1}{2} I

I = \int \frac{dt}{t^{2} + 2 \frac{t}{2} + \frac{1}{4} + a - \frac{1}{4}} = \int \frac{dt}{{(t + \frac{1}{2})}^{2} + \frac{4 a - 1}{4}}

=_{t + \frac{1}{2} = \frac{\sqrt{4 a - 1}}{2} y} \int \frac{1}{(\frac{4 a - 1}{4}) (1 + y^{2})} \frac{\sqrt{4 a - 1}}{2} dy

= \frac{2 \sqrt{4 a - 1}}{4 a - 1} arctany + c = \frac{2}{\sqrt{4 a - 1}} \arctan (\frac{2 t + 1}{\sqrt{4 a - 1}}) + c \Rightarrow

\int \frac{t - a}{t^{2} + t + a} dt = \frac{1}{2} \log (t^{2} + t + a) - \frac{2 a + 1}{\sqrt{4 a - 1}} \arctan (\frac{2 t + 1}{\sqrt{4 a - 1}}) + c \Rightarrow

f (a) = \sqrt{x} + \frac{1}{2} \log (x + \sqrt{x} + a) - \frac{2 a + 1}{\sqrt{4 a - 1}} \arctan (\frac{2 \sqrt{x} + 1}{\sqrt{4 a - 1}}) + c \Rightarrow

f^{^{'}} (a) = \frac{1}{2 (x + \sqrt{x} + a)} - (\frac{2 \sqrt{4 a - 1} - (2 a + 1) \frac{2}{\sqrt{4 a - 1}}}{4 a - 1}) \arctan (\frac{2 \sqrt{x} + 1}{\sqrt{4 a - 1}})

- (\frac{2 a + 1}{\sqrt{4 a - 1}}) \times \frac{(2 \sqrt{x} + 1) {(\frac{1}{\sqrt{4 a - 1}})}^{^{'}}}{1 + {(\frac{2 \sqrt{x} + 1}{\sqrt{4 a - 1}})}^{2}}

= \frac{1}{2 (x + \sqrt{x} + a)} - \frac{2 (4 a - 1) - 2 (2 a + 1)}{(4 a - 1) \sqrt{4 a - 1}} \arctan (\frac{2 \sqrt{x} + 1}{\sqrt{4 a - 1}})

- \frac{2 a + 1}{\sqrt{4 a - 1}} \times \frac{(2 \sqrt{x} + 1) (- \frac{2}{(4 a - 1) \sqrt{4 a - 1}})}{1 + {(\frac{2 \sqrt{x} + 1}{\sqrt{4 a - 1}})}^{2}} \Rightarrow

f^{^{'}} (a) = \frac{1}{2 (x + \sqrt{x} + a)} - \frac{4 a - 4}{(4 a - 1) \sqrt{4 a - 1}} \arctan (\frac{2 \sqrt{x} + 1}{\sqrt{4 a - 1}})

+ \frac{2 (2 a + 1) (2 \sqrt{x} + 1)}{{(4 a - 1)}^{2} (1 + {(\frac{2 \sqrt{x} + 1}{\sqrt{4 a - 1}})}^{2})} \Rightarrow

- f^{^{'}} (1) = - \frac{1}{2 (x + \sqrt{x} + 1)} + \frac{6 (2 \sqrt{x} + 1)}{9 (1 + {(\frac{2 \sqrt{x} + 1}{\sqrt{3}})}^{2}} + K