...nice calculus ... prove : i:∫_0 ^( ∞) ((ln(x))/((1+x^(√2) )^(√2) )) =0 ✓ ii: ∫_0 ^( ∞) (dx/((1+x^(1+(√2)) )^(1+(√2)) )) =(1/( (√2))) ✓ iii: ∫_0 ^( (π/2)) ln(x^2 +ln^2 (cos(x)))dx=πln(ln(2))✓ ... m.n. july.1970...


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Question Number 116014 by mnjuly1970 last updated on 30/Sep/20

$. . . n i c e c a l c u l u s . . .$ $p r o v e :$ $i : \int_{0}^{\infty} \frac{l n (x)}{{(1 + x^{\sqrt{2}})}^{\sqrt{2}}} = 0 ✓$ $i i : \int_{0}^{\infty} \frac{d x}{{(1 + x^{1 + \sqrt{2}})}^{1 + \sqrt{2}}} = \frac{1}{\sqrt{2}} ✓$ $i i i : \int_{0}^{\frac{π}{2}} l n (x^{2} + {l n}^{2} (c o s (x))) d x = π l n (l n (2)) ✓$ $. . . m . n . j u l y . 1970 . . .$
	Answered by mindispower last updated on 30/Sep/20

	$i) t = \frac{1}{x} \Rightarrow$ $i = {\int_{\infty}}^{0} \frac{l n (t) d t}{t^{2} {(\frac{1 + t^{\sqrt{2}}}{t^{\sqrt{2}}})}^{\sqrt{2}}} = - \int_{0}^{\infty} \frac{l n (t) d t}{{(1 + t^{\sqrt{2}})}^{\sqrt{2}}}$ $i = - i \Rightarrow i = 0$ $i i) t = \frac{1}{x}$ $= \int_{0}^{\infty} \frac{t^{1 + 2 \sqrt{2}} d t}{{(t^{1 + \sqrt{2}} + 1)}^{1 + \sqrt{2}}}$ $\int_{0}^{\infty} \frac{t^{1 + 2 \sqrt{2}} + t^{\sqrt{2}}}{{(t^{1 + \sqrt{2}} + 1)}^{1 + \sqrt{2}}} d t - \int_{0}^{\infty} \frac{t^{\sqrt{2}}}{{(t^{1 + \sqrt{2}} + 1)}^{1 + \sqrt{2}}} d t$ $I - J$ $I = \int_{0}^{\infty} \frac{t^{\sqrt{2}}}{{(t^{1 + \sqrt{2}} + 1)}^{\sqrt{2}}} d t$ $t^{1 + \sqrt{2}} = y \Rightarrow d y = (1 + \sqrt{2}) t^{\sqrt{2}}$ $= \int_{0}^{\infty} \frac{d y}{(1 + \sqrt{2}) {(y + 1)}^{\sqrt{2}}} = \int_{0}^{\infty} \frac{1}{(1 + \sqrt{2})} {(y + 1)}^{- \sqrt{2}}$ $= {[- {(y + 1)}^{1 - \sqrt{2}}]}_{0}^{\infty} = 1$ $J, y = t^{1 + \sqrt{2}}$ $d t = \frac{d y}{1 + \sqrt{2}}$ $= \int_{0}^{\infty} \frac{d y}{(1 + \sqrt{2}) {(y + 1)}^{1 + \sqrt{2}}}$ $= \int_{0}^{\infty} \frac{1}{(\sqrt{2} + 1)} {(y + 1)}^{- 1 - \sqrt{2}} d y$ $= [_{0}^{\infty} \frac{{(y + 1)}^{- 1 - \sqrt{2}}}{(1 + \sqrt{2}) (- \sqrt{2})}] = \frac{1}{\sqrt{2} (1 + \sqrt{2})} = \frac{1}{\sqrt{2}} (\sqrt{2} - 1)$ $I - J = 1 - \frac{1}{\sqrt{2}} (\sqrt{2} - 1) = \frac{1}{\sqrt{2}}$ $i i i) n o t n i c e i d e a s$
	Commented by mnjuly1970 last updated on 30/Sep/20

	$g o o d v e r y g o o d$ $y o u r w o r k i s a d m i r a b l e . .$
	Commented by mindispower last updated on 30/Sep/20

	$t h a n k y o u h a v e y o u$ $b o o k f o r i n t e g r a l l i k e t h e t h i r d o n e ?$
	Commented by mnjuly1970 last updated on 30/Sep/20

	$b o o k$ $a d v a n c e d m u r r a y s p i e g e l$ $i s n i c e b o o k$
	Commented by mindispower last updated on 01/Oct/20

	$t h a n k y o u$ $h a v e y o u s o l u t i o n f o r t h e i i i) p l e a s e s i r$
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