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AllQuestion and Answers: Page 759

Question Number 137269 Answers: 1 Comments: 1

Question Number 137255 Answers: 2 Comments: 0

Question Number 137252 Answers: 1 Comments: 0

Decompose the function P(x) = ((x^4 +2x^3 +6x^2 +20x+6)/(x^3 +x^2 +x)) in partial fractions.

$Decompose the function P (x) = \frac{x^{4} + {2 x}^{3} + {6 x}^{2} + 20 x + 6}{x^{3} + x^{2} + x}$ $in partial fractions .$

Question Number 137251 Answers: 2 Comments: 0

......Advanced ... calculus...... 𝛗=∫_0 ^( 1) x^2 ln(x)ln(1−x)dx=???

$. . . . . . A d v a n c e d . . . c a l c u l u s . . . . . .$ $ϕ = \int_{0}^{1} x^{2} l n (x) l n (1 - x) d x = ? ? ?$

Question Number 137249 Answers: 5 Comments: 0

∫ (((3sin x+2))/((2sin x+3)^2 )) dx =?

$\int \frac{(3 \sin x + 2)}{{(2 \sin x + 3)}^{2}} dx = ?$

Question Number 137226 Answers: 1 Comments: 0

Question Number 137225 Answers: 0 Comments: 0

Q137026

$Q 137026$

Question Number 137217 Answers: 0 Comments: 0

Q137014

$Q 137014$

Question Number 137210 Answers: 1 Comments: 0

Question Number 137209 Answers: 1 Comments: 0

Question Number 137208 Answers: 0 Comments: 0

If x,y > 0 then prove that 3(√((x^2 +y^2 )/2)) + (√(xy)) ≥ 2(x+y)

$If x, y > 0 then prove that$ $3 \sqrt{\frac{x^{2} + y^{2}}{2}} + \sqrt{xy} ⩾ 2 (x + y)$

Question Number 137206 Answers: 2 Comments: 0

Question Number 137203 Answers: 2 Comments: 0

∫ ((ln(1+x))/x)=?

$\int \frac{l n (1 + x)}{x} = ?$

Question Number 137197 Answers: 1 Comments: 0

α , β ε (0 , (π/2)) tan^2 α = 1+2tan^2 β ⇒(√2)cosα−cosβ=?

$α, β ϵ (0, \frac{π}{2})$ ${t a n}^{2} α = 1 + 2 {t a n}^{2} β \Rightarrow \sqrt{2} c o s α - c o s β = ?$

Question Number 137192 Answers: 2 Comments: 0

∫ ((cos x)/(1+cos x+sin x)) dx =?

$\int \frac{\cos x}{1 + \cos x + \sin x} dx = ?$

Question Number 137191 Answers: 0 Comments: 0

Some useful approximations of sine function sin((π/7))=((96)/(221)) sin((π/9))=((128)/(373)) sin((π/(11)))=((32)/(113)) ... I am counting more ..thanking you!

$S o m e u s e f u l a p p r o x i m a t i o n s o f s i n e f u n c t i o n$ $s i n (\frac{π}{7}) = \frac{96}{221}$ $s i n (\frac{π}{9}) = \frac{128}{373}$ $s i n (\frac{π}{11}) = \frac{32}{113}$ $. . .$ $I a m c o u n t i n g m o r e . . t h a n k i n g y o u!$

Question Number 137190 Answers: 1 Comments: 0

....advanced ....... calculus..... prove that::: 𝛗=∫_0 ^( 1) ln(ln((1/x))).(dx/( (√(ln((1/x)))))) =−(√π) (γ+ln(4))

$. . . . a d v a n c e d . . . . . . . c a l c u l u s . . . . .$ $p r o v e t h a t :::$ $ϕ = \int_{0}^{1} l n (l n (\frac{1}{x})) . \frac{d x}{\sqrt{l n (\frac{1}{x})}} = - \sqrt{π} (γ + l n (4))$

Question Number 137189 Answers: 0 Comments: 0

Question Number 137187 Answers: 2 Comments: 0

I_n =∫_0 ^(π/2) sin^n x dx Write a relation between I_(n+2) and I_n .

$I_{n} = \int_{0}^{\frac{π}{2}} {s i n}^{n} x d x$ $W r i t e a r e l a t i o n b e t w e e n I_{n + 2} a n d I_{n} .$

Question Number 137185 Answers: 0 Comments: 0

∫_1 ^x e^((1−ln^2 t)^(1/n) ) dt=...? n an integer

$\int_{1}^{x} e^{{(1 - {l n}^{2} t)}^{\frac{1}{n}}} d t = . . . ? n a n i n t e g e r$

Question Number 137177 Answers: 0 Comments: 0

......advanced .... calculus.... Φ=Σ_(k=1) ^∞ ((ψ′(k))/k) =Σ_(n=1) ^∞ (a/n^b ) a , b =?? (adapted from brilliant) ................ ψ(k)=^(??) −γ+∫_0 ^( 1) (((1−t^(k−1) )/(1−t)))dt ∴ ψ′(k)=∫_0 ^( 1) ((−t^(k−1) ln(t))/(1−t))dt Σ_(k=1) ^∞ ((ψ′(k))/k)=∫_0 ^( 1) ((−t^(k−1) ln(t))/((1−t)k))dt =∫_0 ^( 1) ((ln(t))/(t(1−t)))(−(t^k /k))dt=∫_0 ^( 1) ((ln(t)ln(1−t))/(t(1−t)))dt=𝛗 𝛗=∫_0 ^( 1) ((ln(t)ln(1−t))/(t(1−t)))dt=𝛗_1 +𝛗_2 where ... ={∫_0 ^( 1) ((ln(t).ln(1−t))/(1−t))dt=𝛗_1 }+{∫_0 ^( 1) ((ln(1−t).ln(t))/t)dt=𝛗_2 } 𝛗_1 =[−(1/2)ln(t)ln^2 (1−t)]_0 ^1 +(1/2)∫_0 ^( 1) ((ln^2 (1−t))/t)dt ∴ 𝛗_1 = (1/2)(2ζ(3))=ζ(3)=^(easy) 𝛗_2 note : ∫_0 ^( 1) ((ln^2 (1−t))/t)dt=2ζ(3) (derived earlier) 𝛗=𝛗_1 +𝛗_2 =2ζ(3)=Σ_(n=1) ^∞ (2/n^3 ) .... 𝚽= Σ_(n=1) ^∞ (a/n^b ) ......⇒a=2 , b=3

$. . . . . . a d v a n c e d . . . . c a l c u l u s . . . .$ $Φ = \underset{k = 1}{\sum^{\infty}} \frac{ψ^{'} (k)}{k} = \underset{n = 1}{\sum^{\infty}} \frac{a}{n^{b}}$ $a, b = ? ? (a d a p t e d f r o m b r i l l i a n t)$ $. . . . . . . . . . . . . . . .$ $ψ (k) \overset{? ?}{=} - γ + \int_{0}^{1} (\frac{1 - t^{k - 1}}{1 - t}) d t$ $∴ ψ^{'} (k) = \int_{0}^{1} \frac{- t^{k - 1} l n (t)}{1 - t} d t$ $\underset{k = 1}{\sum^{\infty}} \frac{ψ^{'} (k)}{k} = \int_{0}^{1} \frac{- t^{k - 1} l n (t)}{(1 - t) k} d t$ $= \int_{0}^{1} \frac{l n (t)}{t (1 - t)} (- \frac{t^{k}}{k}) d t = \int_{0}^{1} \frac{l n (t) l n (1 - t)}{t (1 - t)} d t = ϕ$ $ϕ = \int_{0}^{1} \frac{l n (t) l n (1 - t)}{t (1 - t)} d t = ϕ_{1} + ϕ_{2} w h e r e . . .$ $= {\int_{0}^{1} \frac{l n (t) . l n (1 - t)}{1 - t} d t = ϕ_{1}} + {\int_{0}^{1} \frac{l n (1 - t) . l n (t)}{t} d t = ϕ_{2}}$ $ϕ_{1} = {[- \frac{1}{2} l n (t) {l n}^{2} (1 - t)]}_{0}^{1} + \frac{1}{2} \int_{0}^{1} \frac{{l n}^{2} (1 - t)}{t} d t$ $∴ ϕ_{1} = \frac{1}{2} (2 ζ (3)) = ζ (3) \overset{e a s y}{=} ϕ_{2}$ $n o t e : \int_{0}^{1} \frac{{l n}^{2} (1 - t)}{t} d t = 2 ζ (3) (d e r i v e d e a r l i e r)$ $ϕ = ϕ_{1} + ϕ_{2} = 2 ζ (3) = \underset{n = 1}{\sum^{\infty}} \frac{2}{n^{3}} . . . .$ $Φ = \underset{n = 1}{\sum^{\infty}} \frac{a}{n^{b}} . . . . . . \Rightarrow a = 2, b = 3$

Question Number 137173 Answers: 1 Comments: 1

Question Number 137171 Answers: 1 Comments: 0

Question Number 137157 Answers: 0 Comments: 6

Question Number 137155 Answers: 2 Comments: 0

......nice calculus ...... evaluate :: 𝛗=∫_0 ^( 2π) (1/(1+cos^4 (x)))dx=???

$. . . . . . n i c e c a l c u l u s . . . . . .$ $e v a l u a t e ::$ $ϕ = \int_{0}^{2 π} \frac{1}{1 + {c o s}^{4} (x)} d x = ? ? ?$

Question Number 137156 Answers: 0 Comments: 0

∫_1 ^∞ ((√(3x^4 +5x^3 +1))/(4x^3 +x^2 +2)) dx = ...

$\underset{1}{\int^{\infty}} \frac{\sqrt{3 x^{4} + 5 x^{3} + 1}}{4 x^{3} + x^{2} + 2} d x = . . .$

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