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IntegrationQuestion and Answers: Page 44

Question Number 166916 Answers: 0 Comments: 0

Ω= Σ_(n=1) ^∞ (( H_( n) )/(n(n+1))) = −−−−−− Ω = Σ_(n=1) ^∞ −(1/(n+1)) ∫_(0 ) ^( 1) x^( n−1) ln(1−x )dx = ∫_0 ^( 1) {−(1/x^2 )ln(1−x).Σ_(n=1) (x^( n+1) /(n+1))}dx = ∫_0 ^( 1) {((−ln(1−x))/x^( 2) )Σ_(n=2) ^∞ (x^( n) /n)}dx = ∫_0 ^( 1) ((−ln(1−x))/x^( 2) ) {−x +Σ_(n=1) ^∞ (x^( n) /n) }dx = −li_( 2) ( 1) +[ ∫_0 ^( 1) ((ln^( 2) ( 1−x ))/x^( 2) )dx=_(derived) ^(earlier) (π^( 2) /3) ] = −(π^( 2) /6) + (π^( 2) /3) = (( π^( 2) )/6) = ζ (2) ■ m.n

$Ω = \underset{n = 1}{\sum^{\infty}} \frac{H_{n}}{n (n + 1)} =$ $- - - - - -$ $Ω = \underset{n = 1}{\sum^{\infty}} - \frac{1}{n + 1} \int_{0}^{1} x^{n - 1} l n (1 - x) d x$ $= \int_{0}^{1} {- \frac{1}{x^{2}} l n (1 - x) . \sum_{n = 1} \frac{x^{n + 1}}{n + 1}} d x$ $= \int_{0}^{1} {\frac{- l n (1 - x)}{x^{2}} \underset{n = 2}{\sum^{\infty}} \frac{x^{n}}{n}} d x$ $= \int_{0}^{1} \frac{- l n (1 - x)}{x^{2}} {- x + \underset{n = 1}{\sum^{\infty}} \frac{x^{n}}{n}} d x$ $= - {l i}_{2} (1) + [\int_{0}^{1} \frac{{l n}^{2} (1 - x)}{x^{2}} d x \underset{d e r i v e d}{\overset{e a r l i e r}{=}} \frac{π^{2}}{3}]$ $= - \frac{π^{2}}{6} + \frac{π^{2}}{3} = \frac{π^{2}}{6} = ζ (2) ◼ m . n$

Question Number 166829 Answers: 1 Comments: 0

calculate If , f(x)=(( (x^2 +1)(((x^( 2) +x−2)(x^( 4) −1)(x^( 2) +2x−3)+16))^(1/3) + (√(x^( 2) +3)))/(( 1+x +x^( 2) ))) then , f ′ (1 ) =? ■ m.n

$c a l c u l a t e$ $I f, f (x) = \frac{(x^{2} + 1) \sqrt[3]{(x^{2} + x - 2) (x^{4} - 1) (x^{2} + 2 x - 3) + 16} + \sqrt{x^{2} + 3}}{(1 + x + x^{2})}$ $t h e n, f^{'} (1) = ? ◼ m . n$

Question Number 166892 Answers: 2 Comments: 0

solve in R i: ⌊ x ⌊ x⌋⌋= 3x ii : ⌊x ⌋^( 2) −3 ⌊x ⌋ +2 ≤ 0 −−−−−−

$s o l v e i n R$ $i : ⌊ x ⌊ x ⌋ ⌋ = 3 x$ $i i : ⌊ x ⌋^{2} - 3 ⌊ x ⌋ + 2 ⩽ 0$ $- - - - - -$

Question Number 166796 Answers: 1 Comments: 0

∫ ((sin^3 x)/((cos^2 x+1)(√(cos^2 x+1)))) dx

$\int \frac{\sin^{3} x}{(\cos^{2} x + 1) \sqrt{\cos^{2} x + 1}} dx$

Question Number 166725 Answers: 2 Comments: 2

∫ (dx/(3+tan x))=?

$\int \frac{dx}{3 + \tan x} = ?$

Question Number 166723 Answers: 1 Comments: 0

∫ (((x)^(1/5) −1)/( (√x) + 1)) dx=?

$\int \frac{\sqrt[5]{x} - 1}{\sqrt{x} + 1} dx = ?$

Question Number 166707 Answers: 1 Comments: 0

(8/(1×5×9))+(8/(5×9×13))+(8/(9×13×17))+…+(1/(41×45×49))=? by M.A

$\frac{8}{1 \times 5 \times 9} + \frac{8}{5 \times 9 \times 13} + \frac{8}{9 \times 13 \times 17} + \dots + \frac{1}{41 \times 45 \times 49} = ?$ $b y M . A$

Question Number 166702 Answers: 0 Comments: 1

∫e^x ln(x)dx=..???

$\int e^{x} l n (x) d x = . . ? ? ?$

Question Number 166684 Answers: 0 Comments: 0

T = ∫ ((sin (x^2 +2))/(2x+4)) dx=?

$T = \int \frac{\sin (x^{2} + 2)}{2 x + 4} dx = ?$

Question Number 166660 Answers: 1 Comments: 0

calculate Ω = Σ_(n=0) ^∞ (1/((3n)!)) = ?

$c a l c u l a t e$ $Ω = \underset{n = 0}{\sum^{\infty}} \frac{1}{(3 n)!} = ?$

Question Number 166653 Answers: 0 Comments: 0

Question Number 166614 Answers: 0 Comments: 0

calculate:: ∫_0 ^∞ (({x}^2 (1−{x})^2 )/((1+x)^5 ))=(7/(12))−γ

$calculate :: \int_{0}^{\infty} \frac{{x}^{2} {(1 - {x})}^{2}}{{(1 + x)}^{5}} = \frac{7}{12} - γ$

Question Number 166610 Answers: 1 Comments: 1

∫ ((x^2 +3)/(x^4 +5x^2 +9)) dx ?

$\int \frac{x^{2} + 3}{x^{4} + {5 x}^{2} + 9} dx ?$

Question Number 166570 Answers: 1 Comments: 0

Question Number 166554 Answers: 0 Comments: 0

∫_(−∞) ^∞ sin ((1/2)πx(x+1)) cos (πx^2 ) dx=?

$\int_{- \infty}^{\infty} \sin (\frac{1}{2} π x (x + 1)) \cos (π x^{2}) dx = ?$

Question Number 166552 Answers: 0 Comments: 0

∫_(−∞) ^∞ cos ((1/2)πx(x+1)) sin (πx^2 ) dx =?

$\int_{- \infty}^{\infty} \cos (\frac{1}{2} π x (x + 1)) \sin (π x^{2}) dx = ?$

Question Number 166491 Answers: 0 Comments: 4

Question Number 166488 Answers: 1 Comments: 0

B=∫ (√((sin 2x−1)/(cos 2x−1))) dx =?

$B = \int \sqrt{\frac{\sin 2 x - 1}{\cos 2 x - 1}} dx = ?$

Question Number 166480 Answers: 0 Comments: 0

Question Number 166449 Answers: 1 Comments: 3

C = ∫ ((1−tan^2 x)/(1+sec^2 x)) dx =?

$C = \int \frac{1 - \tan^{2} x}{1 + \sec^{2} x} dx = ?$

Question Number 166436 Answers: 2 Comments: 0

Show that : Σ_(n=1) ^∞ (((−1)^( n) H_( n) )/n^( 2) ) = −(5/8) ζ (3 ) ■ m.n −−−−−−−−−

$S h o w t h a t :$ $\underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n} H_{n}}{n^{2}} = - \frac{5}{8} ζ (3) ◼ m . n$ $- - - - - - - - -$

Question Number 166376 Answers: 0 Comments: 0

Question Number 166373 Answers: 0 Comments: 0

prove 𝛗=∫_0 ^( 1) (( ln^( 2) (1−x^( 2) ) )/x^( 2) ) dx =(π^( 2) /3) −4ln^( 2) (2) −−− solution (technical method) −−− 𝛗= ∫_0 ^( 1) ln^( 2) (1−x^( 2) )d(1−(1/x)) = [(1−(1/x))ln^( 2) (1−x^( 2) )]_0 ^1 +4∫_0 ^( 1) (1−(1/x))((xln(1−x^( 2) ))/(1−x^( 2) ))dx = −4∫_0 ^( 1) ((ln(1−x^( 2) ))/(1+x)) dx = −4∫_0 ^( 1) ((ln(1+x))/(1+x))dx −4∫_0 ^( 1) ((ln(1−x)dx)/(1+x)) = −2ln^( 2) (2) −4 ( −(π^( 2) /(12)) +(1/2)ln^( 2) (2)) ∴ 𝛗= (π^( 2) /3) −4ln^( 2) (2) ■ m.n

$p r o v e$ $ϕ = \int_{0}^{1} \frac{{l n}^{2} (1 - x^{2})}{x^{2}} d x = \frac{π^{2}}{3} - 4 {l n}^{2} (2)$ $- - - s o l u t i o n (t e c h n i c a l m e t h o d) - - -$ $ϕ = \int_{0}^{1} {l n}^{2} (1 - x^{2}) d (1 - \frac{1}{x})$ $= {[(1 - \frac{1}{x}) {l n}^{2} (1 - x^{2})]}_{0}^{1} + 4 \int_{0}^{1} (1 - \frac{1}{x}) \frac{x l n (1 - x^{2})}{1 - x^{2}} d x$ $= - 4 \int_{0}^{1} \frac{l n (1 - x^{2})}{1 + x} d x$ $= - 4 \int_{0}^{1} \frac{l n (1 + x)}{1 + x} d x - 4 \int_{0}^{1} \frac{l n (1 - x) d x}{1 + x}$ $= - 2 {l n}^{2} (2) - 4 (- \frac{π^{2}}{12} + \frac{1}{2} {l n}^{2} (2))$ $∴ ϕ = \frac{π^{2}}{3} - 4 {l n}^{2} (2) ◼ m . n$

Question Number 166320 Answers: 1 Comments: 0

∫ (dx/(tan^2 x+sin^2 x)) =?

$\int \frac{dx}{\tan^{2} x + \sin^{2} x} = ?$

Question Number 166307 Answers: 1 Comments: 0

∫_0 ^1 ((x^4 (1−x)^4 )/(1+x^2 ))dx

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

Question Number 166263 Answers: 1 Comments: 0

∫_( −(π/4)) ^( (π/4)) (dx/(cos^2 x (√(9+7 tan ∣x∣)))) dx =?

$\int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{dx}{\cos^{2} x \sqrt{9 + 7 \tan ∣ x ∣}} dx = ?$

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