Question-66627

Question Number 66627 by behi83417@gmail.com last updated on 17/Aug/19

Commented by behi83417@gmail.com last updated on 17/Aug/19

You can't use 'macro parameter character #' in math mode

Commented by mathmax by abdo last updated on 18/Aug/19

1) l e t I = \int \sqrt{1 + x + x^{2} + x^{3}} d x \Rightarrow I = \int \sqrt{1 + x + x^{2} (1 + x)} d x

= \int \sqrt{(1 + x) (1 + x^{2})} d x = \int \sqrt{1 + x} \sqrt{1 + x^{2}} d x c h a n g e m e n t \sqrt{1 + x} = t

g i v e 1 + x = t^{2} \Rightarrow x = t^{2} - 1 \Rightarrow I = \int t \sqrt{1 + {(t^{2} - 1)}^{2}} (2 t) d t

= 2 \int t^{2} \sqrt{1 + {(t^{2} - 1)}^{2}} d t l e t t^{2} - 1 = s h (u) \Rightarrow t^{2} = 1 + s h (u) \Rightarrow

2 t d t = c h d u \Rightarrow I = \int \sqrt{1 + s h (u)} c h (u) c h (u) d u

= \int {c h}^{2} (u) \sqrt{1 + s h (u)} d u = \int c h (u) c h (u) (1 + s h {(u)}^{\frac{1}{2}} d u

b y p a r t s f = c h u a n d g^{^{'}} = c h (u) {(1 + s h (u))}^{\frac{1}{2}} \Rightarrow g = \frac{2}{3} {(1 + s h (u))}^{\frac{3}{2}}

\Rightarrow I = \frac{2}{3} c h (u) {(1 + s h (u))}^{\frac{3}{2}} - \frac{2}{3} \int s h (u) {(1 + s h (u))}^{\frac{3}{2}} d u

\dots . b e c o n t i n u e d \dots

Commented by Tanmay chaudhury last updated on 18/Aug/19

c h a l l e n g i n g i n t r e g a l \dots

Commented by Tanmay chaudhury last updated on 18/Aug/19

s i r B e h i p l s u p l o a d t h e s o l u t i o n \dots i f a v a i l a b l e

t h e s e p r o b l e m s s n a t c h e d o u r s l e e p . .

Answered by mind is power last updated on 18/Aug/19

t h e f i r s t {c a n}^{'} t b e e x p r e s s e d u s u a l f u n c t i o n e l e p t i c i n t e g r a l