eplcit-f-x-0-1-ln-x-t-t-2-dt-with-x-gt-1-4-2-calculate-0-1-ln-t-2-t-2-dt-

Question Number 73335 by mathmax by abdo last updated on 10/Nov/19

e p l c i t f (x) = \int_{0}^{1} l n (x + t + t^{2}) d t w i t h x > \frac{1}{4}

2) c a l c u l a t e \int_{0}^{1} l n (t^{2} + t + \sqrt{2}) d t

Commented by mathmax by abdo last updated on 10/Nov/19

1) w e h a v e f (x) = \int_{0}^{1} l n (t^{2} + t + x) d t b y p a r t s w e g e t

f (x) = {[t l n (t^{2} + t + x)]}_{0}^{1} - \int_{0}^{1} t \times \frac{2 t + 1}{t^{2} + t + x} d t

= l n (2 + x) - \int_{0}^{1} \frac{2 t^{2} + t}{t^{2} + t + x} d t w e h a v e

\int_{0}^{1} \frac{2 t^{2} + t}{t^{2} + t + x} d t = \int_{0}^{1} \frac{2 (t^{2} + t + x) - 2 t - 2 x + t}{t^{2} + t + x} d t

= \int_{0}^{1} (2 - \frac{t + 2 x}{t^{2} + t + x}) d t = 2 - \int_{0}^{1} \frac{t + 2 x}{t^{2} + t + x} d t

= 2 - \frac{1}{2} \int_{0}^{1} \frac{2 t + 4 x + 1 - 1}{t^{2} + t + x} d t = 2 - \frac{1}{2} \int_{0}^{1} \frac{2 t + 1}{t^{2} + t + x} d t - \frac{1}{2} \int_{0}^{1} \frac{4 x - 1}{t^{2} + t + x} d t

= 2 - \frac{1}{2} {[l n (t^{2} + t + x)]}_{0}^{1} - \frac{4 x - 1}{2} \int_{0}^{1} \frac{d t}{t^{2} + t + x}

= 2 - \frac{1}{2} (l n (2 + x) - l n (x)) - \frac{4 x - 1}{2} \int_{0}^{1} \frac{d t}{t^{2} + t + x}

\int_{0}^{1} \frac{d t}{t^{2} + t + x} = \int_{0}^{1} \frac{d t}{t^{2} + 2 \frac{t}{2} + \frac{1}{4} + x - \frac{1}{4}} = \int_{0}^{1} \frac{d t}{{(t + \frac{1}{2})}^{2} + \frac{4 x - 1}{4}} (4 x - 1 > 0)

=_{t + \frac{1}{2} = \frac{\sqrt{4 x - 1}}{2} u} \frac{4}{4 x - 1} \int_{\frac{1}{\sqrt{4 x - 1}}}^{\frac{3}{\sqrt{4 x - 1}}} \frac{1}{1 + u^{2}} \frac{\sqrt{4 x - 1}}{2} d u

= \frac{2}{\sqrt{4 x - 1}} {a r c t a n (\frac{3}{\sqrt{4 x - 1}}) - a r c t a n (\frac{1}{\sqrt{4 x - 1}})} \Rightarrow

f (x) = l n (2 + x) - 2 + \frac{1}{2} l n (\frac{2 + x}{x}) + \frac{4 x - 1}{2} \times \frac{2}{\sqrt{4 x - 1}} {a r c t a n (\frac{3}{\sqrt{4 x - 1}}

- a r c t a n (\frac{1}{\sqrt{4 x - 1}})}

f (x) = \frac{3}{2} l n (2 + x) - \frac{1}{2} l n (x) - 2 + \sqrt{4 x - 1} {a r c t a n (\frac{3}{\sqrt{4 x - 1}}) - a r c t a n (\frac{1}{\sqrt{4 x - 1}})}

Commented by mathmax by abdo last updated on 10/Nov/19

2) \int_{0}^{1} l n (t^{2} + t + \sqrt{2}) d t = f (\sqrt{2})

= \frac{3}{2} l n (2 + \sqrt{2}) - \frac{1}{4} l n (2) - 2 + \sqrt{4 \sqrt{2} - 1} {a r c t a n (\frac{3}{\sqrt{4 \sqrt{2} - 1}}) - a r c t a n (\frac{1}{\sqrt{4 \sqrt{2} - 1}})}

Answered by mind is power last updated on 10/Nov/19

f (x) = \int_{0}^{1} \ln (x + t + t^{2}) dt = \ln (x + 2) - \ln (x) - \int \frac{(1 + 2 t) t}{x + t + t^{2}} dt {by part}

= \ln (\frac{x + 2}{2}) - \int_{0}^{1} \frac{{2 t}^{2} + 2 t + 2 x - t - \frac{1}{2} + \frac{1}{2} - 2 x}{t^{2} + t + x} dt

= \ln (\frac{x + 2}{x}) - 2 \int 1 dt + \int \frac{t + \frac{1}{2}}{t^{2} + t + x} + (2 x - \frac{1}{2}) \int_{0}^{1} \frac{dt}{{(t + \frac{1}{2})}^{2} + \frac{4 x - 1}{4}}

= \ln (\frac{x + 2}{x}) - 2 + \frac{1}{2} \ln {(\frac{x + 2}{x})} + \int_{0}^{1} \frac{dt}{{(\frac{2 t}{\sqrt{4 x - 1}} + \frac{1}{\sqrt{4 x - 1}})}^{2} + 1}

= \frac{3}{2} \ln (\frac{x + 2}{x}) - 2 + \frac{\sqrt{4 x - 1}}{2} . {[\arctan (\frac{2 t + 1}{\sqrt{4 x - 1}})]}_{0}^{1}

= \frac{3}{2} \ln (\frac{x + 2}{x}) - 2 + \frac{\sqrt{4 x - 1}}{2} . [arctab (\frac{3}{\sqrt{4 x - 1}}) - \arctan (\frac{1}{\sqrt{4 x - 1}})]

2) x = \sqrt{2}

Commented by mathmax by abdo last updated on 10/Nov/19

t h a n k s s i r .

Commented by mind is power last updated on 10/Nov/19

y^{'} re welcom