find-I-0-1-dx-1-x-1-x-2-

Question Number 30494 by abdo imad last updated on 22/Feb/18

f i n d I = \int_{0}^{1} \frac{d x}{(1 + x) \sqrt{1 + x^{2}}} .

Answered by sma3l2996 last updated on 24/Feb/18

x = s i n h (t) \Rightarrow d x = c o s h (t) d t = \sqrt{1 + {s i n h}^{2} t} d t

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

I = \int_{0}^{l n (\sqrt{2} + 1)} \frac{d t}{1 + s i n h (t)} = \int \frac{e^{t} d t}{e^{2 t} + e^{t} - 1}

u = e^{t} \Rightarrow d u = e^{t} d t

I = \int_{1}^{\sqrt{2} + 1} \frac{d u}{u^{2} + u - 1} = \int \frac{d u}{{(u + \frac{1}{2})}^{2} - \frac{5}{4}}

I = \frac{4}{5} \int_{1}^{\sqrt{2} + 1} \frac{d u}{{(\frac{2}{\sqrt{5}} \times \frac{2 u + 1}{2})}^{2} - 1} = \frac{4}{5} \int \frac{d u}{{(\frac{2 u + 1}{\sqrt{5}})}^{2} - 1}

y = \frac{2 u + 1}{\sqrt{5}} \Rightarrow d u = \frac{\sqrt{5}}{2} d y

I = \frac{2 \sqrt{5}}{5} \int_{3 / \sqrt{5}}^{(2 \sqrt{2} + 3) / \sqrt{5}} \frac{d y}{y^{2} - 1} = \frac{2 \sqrt{5}}{5} \int \frac{d y}{(y + 1) (y - 1)}

I = \frac{\sqrt{5}}{5} \int_{3 / \sqrt{5}}^{(2 \sqrt{2} + 3) / \sqrt{5}} (\frac{1}{y - 1} - \frac{1}{y + 1}) d y = \frac{\sqrt{5}}{5} {[l n ∣ \frac{y - 1}{y + 1} ∣]}_{3 / \sqrt{5}}^{(2 \sqrt{2} + 3) / \sqrt{5}}