Question-94857

Question Number 94857 by i jagooll last updated on 21/May/20

Commented by prakash jain last updated on 21/May/20

Can you write without bold font ?

It looks too dark .

Commented by i jagooll last updated on 21/May/20

on my cellphone it is not dark

Commented by i jagooll last updated on 21/May/20

and that's the result of the photo

Commented by mr W last updated on 21/May/20

y o u s h o u l d t r y t o c r o p y o u r i m a g e

p r o p e r l y, i . e . t o k e e p m o r e w r i t e

s p a c e a r o u n d t h e f o r m u l a (i n t h i s

c a s e m o r e w r i t e s p a c e o n l e f t a n d r i g h t

s i d e f r o m t h e f o r m u l a), t h e n y o u

w i l l g e t p r o p e r l y s i z e d i m a g e a f t e r

p o s t i n g . c u r r e n t l y t h e i m a g e s y o u

p o s t e d l o o k v e r y t i n y .

t h i s i s a n e x a m p l e w h a t i d o :

t a p y o u r i m a g e a n d m y i m a g e, y o u

w i l l s e e w h e r e t h e d i f f e r e n c e i s .

Commented by mr W last updated on 21/May/20

Commented by mr W last updated on 21/May/20

i t h i n k i k n o w h o w t h e p r o b l e m c o m e s .

y o u w r i t e t h e f o r m u l a w i t h t h e e d i t o r

o f t h i s a p p a n d t h e n s a v e i t a s i m a g e

a n d t h e n p o s t i t i n t o t h e f o r u m . b u t

t h e e d i t o r s a v e s t h e f o r m u l a a s i m a g e

w i t h o u t a n y e x t r a w r i t e s p a c e a r o u n d

t h e f o r m u l a . i f y o u {d o n}^{'} t t r e a t s u c h

a n i m a g e w i t h a n o t h e r a p p, y o u

{w o n}^{'} t b e a b l e t o s o l v e t h i s p r o b l e m .

b u t m y q u e s t i o n : w h y {d o n}^{'} t y o u w r i t e

t h e f o r m u l a d i r e c t l y i n t h e f o r u m ?

Commented by Tinku Tara last updated on 21/May/20

We reverted changes so that images

will again be scaled to width . So

they will look big . Please update

to latest version . 2.71 .

We are still working on a proper

solution for very small images .

Thank You .

Answered by mathmax by abdo last updated on 21/May/20

I = \int \frac{{(x + 1)}^{2}}{x^{2} {(x^{2} + 1)}^{2}} dx complex method let decompose

F (x) = \frac{x^{2} + 2 x + 1}{x^{2} {(x^{2} + 1)}^{2}} \Rightarrow F (x) = \frac{x^{2} + 2 x + 1}{x^{2} {(x - i)}^{2} {(x + i)}^{2}}

F (x) = \frac{a}{x} + \frac{b}{x^{2}} + \frac{c}{x - i} + \frac{d}{{(x - i)}^{2}} + \frac{e}{x + i} + \frac{f}{{(x + i)}^{2}}

b = 1, d = \frac{2 i}{- {(2 i)}^{2}} = \frac{- 2 i}{- 4} = \frac{i}{2}

f = \frac{- 2 i}{- {(- 2 i)}^{2}} = \frac{2 i}{- 4} = - \frac{i}{2} \Rightarrow F (x) = \frac{a}{x} + \frac{1}{x^{2}} + \frac{c}{x - i} + \frac{i}{2 {(x - i)}^{2}}

+ \frac{e}{x + i} - \frac{i}{2 {(x + i)}^{2}}

\lim_{x \to + \infty} xF (x) = 0 = a + c + e

F (1) = 1 = a + 1 + \frac{c}{1 - i} + \frac{i}{2 {(1 - i)}^{2}} + \frac{e}{1 + i} - \frac{i}{2 {(1 + i)}^{2}}

= a + 1 + (1 + i) \frac{c}{2} + \frac{i}{2 (- 2 i)} + (1 - i) \frac{e}{2} - \frac{i}{2 (2 i)} \Rightarrow

a + \frac{1}{2} + \frac{1}{2} (1 + i) c + \frac{1}{2} (1 - i) e = 1

after we calculate F (- 1) we get the value of a c ande \Rightarrow

\int F (x) dx = aln ∣ x ∣ - \frac{1}{x} + cln (x - i) - \frac{i}{2 (x - i)} + eln (x + i) + \frac{i}{2 (x + i)} + C

= aln ∣ x ∣ - \frac{1}{x} + cln (x - i) + eln (x + i) + \frac{i}{2} (\frac{1}{x + i} - \frac{1}{x - i}) + C

= aln ∣ x ∣ - \frac{1}{x} + cln (x - i) + eln (x + i) + \frac{1}{x^{2} + 1} + C

Answered by niroj last updated on 21/May/20

I = \int \frac{x^{2} + 2 x + 1}{{x (x^{2} + 1)}^{2}} dx

= \int \frac{x^{2} + 1 + 2 x}{x^{2} {(x^{2} + 1)}^{2}} dx

= \int \frac{(x^{2} + 1)}{x^{2} {(x^{2} + 1)}^{2}} dx + \int \frac{2 x}{x^{2} {(x^{2} + 1)}^{2}} dx

= \int \frac{1}{x^{2} (x^{2} + 1)} dx + \int \frac{2 x}{x^{2} (x^{2} + 1)} dx

I = I_{1} + I_{2}

For, I_{2}

Put, x^{2} + 1 = t, x^{2} = t - 1

2 xdx = dt

\int \frac{1}{x^{2}} dx - \int \frac{1}{x^{2} + 1} dx + \int \frac{1}{(t - 1) t} dt

[\frac{x^{- 2 + 1}}{- 2 + 1}] - \tan^{- 1} x + \int \frac{1}{t - 1} dt - \int \frac{1}{t} dt + C

\frac{x^{- 1}}{- 1} - \tan^{- 1} x + \log (t - 1) - \log t + C

- \frac{1}{x} - \tan^{- 1} x + \log \frac{t - 1}{t} + C

- \frac{1}{x} - \tan^{- 1} x + \log \frac{x^{2} + 1 - 1}{x^{2} + 1} + C

- \frac{1}{x} - \tan^{- 1} x + \log \frac{x^{2}}{x^{2} + 1} + C / / .

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