Evaluate-0-1-x-4-1-x-4-1-x-2-dx-

Question Number 17463 by alex041103 last updated on 06/Jul/17

E v a l u a t e \underset{0}{\int^{1}} \frac{x^{4} {(1 - x)}^{4}}{1 + x^{2}} d x .

Answered by ajfour last updated on 06/Jul/17

\frac{22}{7} - π .

Commented by alex041103 last updated on 06/Jul/17

T r u e . {I s n}^{'} t t h a t b e a u t i f u l l . (Sorry,

if I made a mistake somewhere . I^{'} m

only 13 years old .)

Commented by ajfour last updated on 06/Jul/17

excellent expression!

Answered by alex041103 last updated on 17/Jul/17

F i r s t

{(1 - x)}^{4} = 1 - 4 x + 6 x^{2} - 4 x^{3} + x^{4}

W e c a n n o w d o l o n g d i v i s i o n

x^{4} \frac{{(1 - x)}^{4}}{1 + x^{2}} = x^{4} (x^{2} - 4 x + 5 - \frac{4}{1 + x^{2}}) =

= x^{6} - 4 x^{5} + 5 x^{4} - 4 \frac{x^{4}}{1 + x^{2}} = (a g a i n l o n g d i v i s i o n)

= x^{6} - 4 x^{5} + 5 x^{4} - 4 x^{2} + 4 - 4 \frac{1}{1 + x^{2}}

A n d n o w w e i n t e g r a t e a n d e v a l u a t e

\Rightarrow \int_{0}^{1} \frac{x^{4} {(1 - x)}^{4}}{1 + x^{2}} d x = \frac{22}{7} - π

n o t e :

4 \int_{0}^{1} \frac{1}{1 + x^{2}} d x = 4 {[{t a n}^{- 1} (x)]}_{0}^{1} = 4 \frac{π}{4} = π

Commented by alex041103 last updated on 18/Jul/17

d i v i d i o n o f p o l i n o m i a l s

Commented by alex041103 last updated on 19/Jul/17

{Y o u}^{'} l l d i n d s o m e i n f o a t

goo . gl / Qo 7 V 6 H