find-x-1-1-x-2-1-x-2-x-1-dx-

Question Number 40620 by math khazana by abdo last updated on 25/Jul/18

f i n d \int ((x + 1) \sqrt{1 + x^{2}} + (1 + x^{2}) \sqrt{x + 1}) d x

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Jul/18

\int x \sqrt{1 + x^{2}} d x + \int \sqrt{x^{2} + 1} d x + \int \sqrt{x + 1} d x + \int x^{2} \sqrt{x + 1}

I_{1} = \int x \sqrt{1 + x^{2}} d x

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

= \frac{1}{2} \times \frac{2}{3} {(1 + x^{2})}^{\frac{3}{2}}

I_{2} = \int \sqrt{x^{2} + 1} d x u s e f o r m u l a

= \frac{x}{2} \sqrt{x^{2} + 1} + \frac{1}{2} l n (x + \sqrt{x^{2} + 1})

I_{3} = \int \sqrt{x + 1} d x

= \frac{{(x + 1)}^{\frac{3}{2}}}{\frac{3}{2}}

I_{4} = \int x^{2} \sqrt{x + 1} d x

l e t t^{2} = x + 1 2 t d t = d x

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

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

2 \int t^{6} - 2 t^{4} + t^{2} d t

2 (\frac{t^{7}}{7} - \frac{2 t^{5}}{5} + \frac{t^{3}}{3})

\frac{2}{7} \frac{t^{8}}{8} - \frac{4}{5} \times \frac{t^{6}}{6} + \frac{2}{3} \times \frac{t^{4}}{4}

\frac{}{}

Answered by MJS last updated on 25/Jul/18

\int x^{2} \sqrt{x + 1} d x = \frac{2}{7} (x^{2} - \frac{4 x}{5} + \frac{8}{15}) {(x + 1)}^{\frac{3}{2}}

[\begin{matrix} t = x + 1 \to d x = d t \\ \int {(t - 1)}^{2} \sqrt{t} d t = \int (t^{2} - 2 t + 1) \sqrt{t} d t = \int (t^{\frac{5}{2}} - 2 t^{\frac{3}{2}} + t^{\frac{1}{2}}) d t \\ \int t^{q} d t = \frac{1}{q + 1} t^{q + 1} \end{matrix}]

\int x \sqrt{x^{2} + 1} d x = \frac{1}{3} {(x^{2} + 1)}^{\frac{3}{2}}

[\begin{matrix} t = x^{2} + 1 \to d x = \frac{d t}{2 x} \\ \frac{1}{2} \int \sqrt{t} d t \\ \int t^{q} d t = \frac{1}{q + 1} t^{q + 1} \end{matrix}]

\int \sqrt{x^{2} + 1} d x = \frac{1}{2} (x \sqrt{x^{2} + 1} + \ln ∣ x + \sqrt{x^{2} + 1} ∣)

[\begin{matrix} t = \arctan x \to d x = \sec^{2} t d t \\ \int \sec^{2} t \sqrt{1 + \tan^{2} t} d t = \int \sec^{3} t d t \\ \int \sec^{n} t d t = \frac{\sec^{n - 2} t \tan t}{n - 1} + \frac{n - 2}{n - 1} \int \sec^{n - 2} t d t \end{matrix}]

\int \sqrt{x + 1} d x = \frac{2}{3} {(x + 1)}^{\frac{3}{2}}

[\int {(x + a)}^{q} d x = \frac{1}{q + 1} {(x + a)}^{q + 1}]

Answered by maxmathsup by imad last updated on 25/Jul/18

I = \int (x + 1) \sqrt{1 + x^{2}} d x + \int (1 + x^{2}) \sqrt{x + 1} = K + H

K = \int (x + 1) \sqrt{1 + x^{2}} d x =_{x = s h (t)} \int (s h (t) + 1) c h (t) c h (t) d x

= \int (s h (t) + 1) {c h}^{2} t d t = \int s h (t) {c h}^{2} (t) d t + \int {c h}^{2} t d t

= \frac{1}{3} {c h}^{3} (t) + \int \frac{1 + c h (2 t)}{2} d t

= \frac{1}{3} {c h}^{3} (t) + \frac{t}{2} + \frac{1}{4} s h (2 t) = \frac{1}{3} {\frac{e^{t} + e^{- t}}{2}} + \frac{t}{2} + \frac{1}{2} c h (t) s h (t)

= \frac{1}{6} {x + \sqrt{1 + x^{2}} + {(x + \sqrt{1 + x^{2}})}^{- 1}} + \frac{l n (x + \sqrt{1 + x^{2})}}{2} + \frac{x \sqrt{1 + x^{2}}}{2} + c_{1}

H = \int (1 + x^{2}) \sqrt{x + 1} d x =_{\sqrt{x + 1} = t} \int (1 + {(t^{2} - 1)}^{2}) t (2 t d t)

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

= 2 \int (t^{6} - 2 t^{4} + 2 t^{2}) d t = 2 {\frac{t^{7}}{7} - \frac{2}{5} t^{5} + \frac{2}{3} t^{3}} + c_{2}

= \frac{2}{7} {(\sqrt{x + 1})}^{7} - \frac{4}{5} {(\sqrt{x + 1})}^{5} + \frac{4}{3} {(\sqrt{1 + x})}^{3} + c_{2}

I = K + H t h e v a l u e o f I i s d e t e r m i n e d .

Commented by maxmathsup by imad last updated on 25/Jul/18

e r r o r a t l i n e 5

K = \frac{1}{3} {c h}^{3} (t) + \frac{t}{2} s h (2 t) = \frac{1}{3} {\frac{e^{t} + e^{- t}}{2}}^{3} + \frac{t}{2} + \frac{1}{2} c h (t) s h (t)

= \frac{1}{24} {x + \sqrt{1 + x^{2}} + {(x + \sqrt{1 + x^{2}})}^{- 1}}^{3} + \frac{l n (x + \sqrt{1 + x^{2})}}{2} + \frac{x \sqrt{1 + x^{2}}}{2} + c_{1}