evaluate-2-0-2-x-1-x-2-4-dx-

Question Number 83675 by niroj last updated on 05/Mar/20

evaluate :

2 \int_{0}^{2} \frac{\sqrt{x + 1}}{x^{2} + 4} dx

Commented by jagoll last updated on 05/Mar/20

\sqrt{x + 1} = u \Rightarrow x^{2} = {(u^{2} - 1)}^{2}

2 x d x = 4 u (u^{2} - 1) d u

d x = \frac{2 u (u^{2} - 1) d u}{u^{2} - 1} = 2 u d u

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

2 {\int^{\sqrt{3}}}_{1} \frac{u (2 u d u)}{4 + {(u^{2} - 1)}^{2}}

l e t u = \sec t \Rightarrow du = \sec t \tan t d t

Commented by niroj last updated on 05/Mar/20

t h a n k s m r . j a g o l l t r y t o s o l v e i f i t i s b e t t e r t o b e c o m p l e t e .

Answered by TANMAY PANACEA last updated on 05/Mar/20

I = \int \frac{x + 1}{\sqrt{x + 1} \times (x^{2} + 4)} d x

t^{2} = x + 1

\int \frac{t^{2} \times 2 t d t}{t (t^{4} - 2 t^{2} + 1 + 4)}

2 \int \frac{t^{2}}{t^{4} - 2 t^{2} + 5} d t

2 \int \frac{d t}{t^{2} + \frac{5}{t^{2}} - 2}

t^{2} + \frac{5}{t^{2}} = {(t + \frac{\sqrt{5}}{t})}^{2} - 2 \sqrt{5}

\frac{d}{d t} (t + \frac{\sqrt{5}}{t}) = 1 - \frac{\sqrt{5}}{t^{2}}

t^{2} + \frac{5}{t^{2}} = {(t - \frac{\sqrt{5}}{t})}^{2} + 2 \sqrt{5}

\frac{d}{d t} (t - \frac{\sqrt{5}}{t}) = 1 + \frac{\sqrt{5}}{t^{2}}

l o o k (1 + \frac{\sqrt{5}}{t^{2}}) + (1 - \frac{\sqrt{5}}{t^{2}}) = 2

n o w \int \frac{1 - \frac{\sqrt{5}}{t^{2}} + 1 + \frac{\sqrt{5}}{t^{2}}}{t^{2} + \frac{5}{t^{2}} - 2} d t

\int \frac{d (t + \frac{\sqrt{5}}{t})}{{(t + \frac{\sqrt{5}}{t})}^{2} - 2 \sqrt{5} - 2} + \int \frac{d (t - \frac{\sqrt{5}}{t})}{{(t - \frac{\sqrt{5}}{t})}^{2} + 2 \sqrt{5} - 2}

n o w p l s u s e f o r m u l a o f

\int \frac{d x}{x^{2} - a^{2}} a n d \int \frac{d x}{x^{2} + a^{2}}

2 \sqrt{5} + 2 = {(\sqrt{2 \sqrt{5} + 2})}^{2} \leftarrow a s i f a^{2}

2 \sqrt{5} - 2 = {(\sqrt{2 \sqrt{5} - 2})}^{2}

s o

\frac{1}{2 (\sqrt{2 \sqrt{5} + 2})} l n (\frac{(t + \frac{\sqrt{5}}{t}) - (\sqrt{2 \sqrt{5} + 2})}{(t + \frac{\sqrt{5}}{t}) + (\sqrt{2 \sqrt{5} + 2})}) + \frac{1}{\sqrt{2 \sqrt{5} - 2}} {t a n}^{- 1} (\frac{t - \frac{\sqrt{5}}{t}}{\sqrt{2 \sqrt{5} - 2}})

n o w p l s r e p l a c e t b y \sqrt{x + 1} a n d p u t u p p e r

a n d l o w e r l i m i t

Commented by niroj last updated on 05/Mar/20

y o u a r e a b s u t e a m a z i n g . n i c e s o l u t i o n . .

Commented by TANMAY PANACEA last updated on 05/Mar/20

t h a n k y o u s i r