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Question Number 138299 Answers: 2 Comments: 0

Question Number 138296 Answers: 3 Comments: 0

∫ (dx/(x^4 (√(x^2 −a^2 )))) =?

$\int \frac{d x}{x^{4} \sqrt{x^{2} - a^{2}}} = ?$

Question Number 138295 Answers: 1 Comments: 0

hi ! Simplify : for n ∈ N^∗ , S_n = Σ_(k=1) ^n ((3k+8)/(k(k+2)2^k )) .

$hi!$ $Simplify : f o r n \in N^{*}, S_{n} = \underset{k = 1}{\sum^{n}} \frac{3 k + 8}{k (k + 2) 2^{k}} .$

Question Number 138283 Answers: 0 Comments: 2

......advanced ........... calculus...... prove that:: 𝛗=∫_0 ^( 1) ((ln(1+x^2 ).arctan(x))/x^2 )dx= proof::: 𝛗=_(⟨substitution⟩) ^(x=tan(θ)) ∫_0 ^( (π/4)) ((ln(1+tan^2 (θ)).θ)/(tan^2 (θ)))(1+tan^2 (θ))dθ =^(⟨simplification⟩) ∫_0 ^( (π/4)) ((θ.ln((1/(cos^2 (θ)))))/(sin^2 (θ)))dθ =−2∫_0 ^( (π/4)) ((θ.ln(cos(θ))/(sin^2 (θ)))dθ =^(i.b.p) 2{[(cot(θ).θ.ln(cos(θ))]_0 ^(π/4) −∫_0 ^( (π/4)) (cot(θ).[ln(cos(θ))−θ.tan(θ)]dθ =2.(π/4).ln(((√2)/2))−2∫_0 ^( (π/4)) cot(θ).ln(cos(θ))dθ+2∫_0 ^( (π/4)) θdθ =((−π)/4)ln(2)−Φ+(π^2 /(16)) Φ=∫_0 ^( (π/4)) ((cos(θ))/(sin(θ))).ln(1−sin^2 (θ))dθ =^(sin(θ)=y) ∫_0 ^( ((√2)/2)) ((ln(1−y^2 ))/y)dy=−∫_0 ^( ((√2)/2)) Σ_(n=1) ^∞ (y^(2n−1) /n)dy =−Σ[(y^(2n) /(2n^2 ))]_0 ^( ((√2)/2)) =((−1)/2) li_2 ((1/2)) =((−1)/2){(π^2 /(12))−(1/2)ln^2 (2)}=((−π^2 )/(24))+(1/4)ln^2 (2)... ∴ 𝛗=((−π)/4)ln(2)+(π^2 /(24))+(1/4)ln^2 (2)+(π^2 /(16)) ......... 𝛗 =((5π^2 )/(48))−((12π)/(48))ln(2)+((12)/(48)) ln^2 (2) ..... ........𝛗=(1/(48)){5π^2 −12πln(2)+12ln^2 (2)}

$. . . . . . a d v a n c e d . . . . . . . . . . . c a l c u l u s . . . . . .$ $p r o v e t h a t ::$ $ϕ = \int_{0}^{1} \frac{l n (1 + x^{2}) . a r c t a n (x)}{x^{2}} d x =$ $p r o o f :::$ $ϕ \underset{⟨ s u b s t i t u t i o n ⟩}{\overset{x = t a n (θ)}{=}} \int_{0}^{\frac{π}{4}} \frac{l n (1 + {t a n}^{2} (θ)) . θ}{{t a n}^{2} (θ)} (1 + {t a n}^{2} (θ)) d θ$ $\overset{⟨ s i m p l i f i c a t i o n ⟩}{=} \int_{0}^{\frac{π}{4}} \frac{θ . l n (\frac{1}{{c o s}^{2} (θ)})}{{s i n}^{2} (θ)} d θ$ $= - 2 \int_{0}^{\frac{π}{4}} \frac{θ . l n (c o s (θ)}{{s i n}^{2} (θ)} d θ$ $\overset{i . b . p}{=} 2 {[{(c o t (θ) . θ . l n (c o s (θ))]}_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} (c o t (θ) . [l n (c o s (θ)) - θ . t a n (θ)] d θ$ $= 2 . \frac{π}{4} . l n (\frac{\sqrt{2}}{2}) - 2 \int_{0}^{\frac{π}{4}} c o t (θ) . l n (c o s (θ)) d θ + 2 \int_{0}^{\frac{π}{4}} θ d θ$ $= \frac{- π}{4} l n (2) - Φ + \frac{π^{2}}{16}$ $Φ = \int_{0}^{\frac{π}{4}} \frac{c o s (θ)}{s i n (θ)} . l n (1 - {s i n}^{2} (θ)) d θ$ $\overset{s i n (θ) = y}{=} \int_{0}^{\frac{\sqrt{2}}{2}} \frac{l n (1 - y^{2})}{y} d y = - \int_{0}^{\frac{\sqrt{2}}{2}} \underset{n = 1}{\sum^{\infty}} \frac{y^{2 n - 1}}{n} d y$ $= - Σ {[\frac{y^{2 n}}{2 n^{2}}]}_{0}^{\frac{\sqrt{2}}{2}} = \frac{- 1}{2} {l i}_{2} (\frac{1}{2})$ $= \frac{- 1}{2} {\frac{π^{2}}{12} - \frac{1}{2} {l n}^{2} (2)} = \frac{- π^{2}}{24} + \frac{1}{4} {l n}^{2} (2) . . .$ $∴ ϕ = \frac{- π}{4} l n (2) + \frac{π^{2}}{24} + \frac{1}{4} {l n}^{2} (2) + \frac{π^{2}}{16}$ $. . . . . . . . . ϕ = \frac{5 π^{2}}{48} - \frac{12 π}{48} l n (2) + \frac{12}{48} {l n}^{2} (2) . . . . .$ $. . . . . . . . ϕ = \frac{1}{48} {5 π^{2} - 12 π l n (2) + 12 {l n}^{2} (2)}$

Question Number 138280 Answers: 0 Comments: 2

x^(x+4) =32 solution method?

$x^{x + 4} = 32 solution method ?$

Question Number 138278 Answers: 1 Comments: 0

Question Number 138277 Answers: 3 Comments: 0

calculate ∫_0 ^∞ ∫_0 ^∞ e^(−x^2 −y^2 ) sin(x^2 +y^2 )dxdy

$calculate \int_{0}^{\infty} \int_{0}^{\infty} e^{- x^{2} - y^{2}} \sin (x^{2} + y^{2}) dxdy$

Question Number 138275 Answers: 0 Comments: 0

1)calculate U_n =∫∫_([(1/n),1]^2 ) (2x+3y)(√(x^2 +y^2 ))dxdy 2)find ∫∫_(]0,1]^2 ) (2x+3y)(√(x^2 +y^2 ))dxdy

$1) calculate U_{n} = \int \int_{{[\frac{1}{n}, 1]}^{2}} (2 x + 3 y) \sqrt{x^{2} + y^{2}} dxdy$ $2) find \int \int_{] {0, 1]}^{2}} (2 x + 3 y) \sqrt{x^{2} + y^{2}} dxdy$

Question Number 138276 Answers: 0 Comments: 0

1) calculate A_n =∫∫_([0,n[^2 ) ((dxdy)/((2x^2 +3y^2 )^2 )) 2)find lim_(n→+∞) A_n

$1) calculate A_{n} = \int \int_{[0, n [^{2}} \frac{dxdy}{{({2 x}^{2} + {3 y}^{2})}^{2}}$ $2) find \lim_{n \to + \infty} A_{n}$

Question Number 138240 Answers: 3 Comments: 0

(x−1)(dy/dx) +xy = 2xe^(−x)

$(x - 1) \frac{d y}{d x} + x y = 2 {x e}^{- x}$

Question Number 138257 Answers: 1 Comments: 0

hi ! calculate : ∫∫_A (x^2 −y^2 )dxdy with A={(x^2 /a^2 ) + (y^2 /b^2 ) ≤ 1}

$hi!$ $calculate :$ $\int \int_{A} (x^{2} - y^{2}) d x d y w i t h A = {\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} ⩽ 1}$

Question Number 138254 Answers: 0 Comments: 10

Question Number 138250 Answers: 1 Comments: 0

(x^2 −10x+6)^(x−2) >1

${(x^{2} - 10 x + 6)}^{x - 2} > 1$

Question Number 138249 Answers: 1 Comments: 0

lim_(n→0) ((2^(n+1) +3^(n+2) +4^(n+3) )/(2^n +3^n +4^n ))

$li \underset{n \to 0}{m} \frac{2^{n + 1} + 3^{n + 2} + 4^{n + 3}}{2^{n} + 3^{n} + 4^{n}}$

Question Number 138248 Answers: 0 Comments: 3

x^(x+4) =32

$x^{x + 4} = 32$

Question Number 138236 Answers: 0 Comments: 5

prove that (e^(−2y) /(1+e^(−2y) ))<∣tan(z) − i∣<(e^(−2y) /(1−e^(−2y) )) ,y>0 how can solve this ?

$p r o v e t h a t$ $\frac{e^{- 2 y}}{1 + e^{- 2 y}} <∣ t a n (z) - i ∣< \frac{e^{- 2 y}}{1 - e^{- 2 y}}, y > 0$ $h o w c a n s o l v e t h i s ?$

Question Number 138235 Answers: 1 Comments: 0

for p,q∈R satisfying p^4 +q^4 =4pq find the range of p+q when 1) no restriction 2) 0≤p≤1, 0≤q≤1

$f o r p, q \in R s a t i s f y i n g p^{4} + q^{4} = 4 p q$ $f i n d t h e r a n g e o f p + q w h e n$ $1) n o r e s t r i c t i o n$ $2) 0 ⩽ p ⩽ 1, 0 ⩽ q ⩽ 1$

Question Number 138231 Answers: 0 Comments: 0

Question Number 138224 Answers: 1 Comments: 1

Question Number 138223 Answers: 2 Comments: 0

.......nice ... ... ... calculus... evaluate ::: 𝚯=Σ_(n=−∞) ^∞ (1/((3n+1)^3 )) =? .........................

$. . . . . . . n i c e . . . . . . . . . c a l c u l u s . . .$ $e v a l u a t e :::$ $Θ = \underset{n = - \infty}{\sum^{\infty}} \frac{1}{{(3 n + 1)}^{3}} = ?$ $. . . . . . . . . . . . . . . . . . . . . . . . .$

Question Number 138219 Answers: 2 Comments: 0

Consider the function y=((x^2 +3x+6)/(3x−1)) 1) Determine the intercept of the function 2) Find the asymptotes if they exist 3) find the turning point and determine the type of turnin point they are. 4) sketch the graph of the function.

$Consider the function y = \frac{x^{2} + 3 x + 6}{3 x - 1}$ $1) Determine the intercept of the function$ $2) Find the asymptotes if they exist$ $3) find the turning point and$ $determine the type of turnin point they$ $are .$ $4) sketch the graph of the function .$

Question Number 138209 Answers: 2 Comments: 1

I_n =∫_0 ^( 1) x^n (√(x^2 +1))dx find reduction formula

$I_{n} = \int_{0}^{1} x^{n} \sqrt{x^{2} + 1} d x$ $f i n d r e d u c t i o n f o r m u l a$

Question Number 138206 Answers: 1 Comments: 1

Question Number 138205 Answers: 2 Comments: 3

Question Number 138203 Answers: 1 Comments: 0

Three circles each radius 1, touch one another externally and they lie between two parallel line. The minimum possible distance between the lines is _

$T h r e e c i r c l e s e a c h r a d i u s 1, t o u c h o n e$ $a n o t h e r e x t e r n a l l y a n d t h e y l i e$ $b e t w e e n t w o p a r a l l e l l i n e . T h e$ $m i n i m u m p o s s i b l e d i s t a n c e b e t w e e n$ $t h e l i n e s i s_$

Question Number 142191 Answers: 1 Comments: 3

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