find ∫ ((x+1)(√(1+x^2 )) +(1+x^2 )(√(x+1)))dx


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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
	$I = ∫ (x+1)(√(1+x^2 ))dx +∫(1+x^2 )(√(x+1))=K+H K = ∫ (x+1)(√(1+x^2 ))dx=_(x=sh(t)) ∫ (sh(t)+1)ch(t)ch(t)dx = ∫ (sh(t)+1)ch^2 t dt = ∫ sh(t)ch^2 (t)dt +∫ ch^2 t dt =(1/3)ch^3 (t) + ∫ ((1+ch(2t))/2)dt =(1/3)ch^3 (t) +(t/2) +(1/4)sh(2t) =(1/3){((e^t +e^(−t) )/2)} +(t/2) +(1/2)ch(t)sh(t) =(1/6){x+(√(1+x^2 )) + (x+(√(1+x^2 )))^(−1) } +((ln(x+(√(1+x^2 ))))/2)+((x(√(1+x^2 )))/2) +c_1 H = ∫ (1+x^2 )(√(x+1))dx =_((√(x+1)) =t) ∫ (1+(t^2 −1)^2 )t (2tdt) =2 ∫ t^2 (1+t^4 −2t^2 +1)dt =2 ∫ t^2 (t^4 −2t^2 +2)dt =2 ∫ (t^6 −2t^4 +2t^2 )dt =2{(t^7 /7) −(2/5)t^5 +(2/3)t^3 } +c_2 =(2/7)((√(x+1)))^7 −(4/5) ((√(x+1)))^(5 ) +(4/3) ((√(1+x)))^3 +c_2 I =K +H the value of I is determined.$
	$I = \int (x + 1) \sqrt{1 + x^{2}} d x + \int (1 + x^{2}) \sqrt{x + 1} = K + H$ $K = \int (x + 1) \sqrt{1 + x^{2}} d x =_{x = s h (t)} \int (s h (t) + 1) c h (t) c h (t) d x$ $= \int (s h (t) + 1) {c h}^{2} t d t = \int s h (t) {c h}^{2} (t) d t + \int {c h}^{2} t d t$ $= \frac{1}{3} {c h}^{3} (t) + \int \frac{1 + c h (2 t)}{2} d t$ $= \frac{1}{3} {c h}^{3} (t) + \frac{t}{2} + \frac{1}{4} s h (2 t) = \frac{1}{3} {\frac{e^{t} + e^{- t}}{2}} + \frac{t}{2} + \frac{1}{2} c h (t) s h (t)$ $= \frac{1}{6} {x + \sqrt{1 + x^{2}} + {(x + \sqrt{1 + x^{2}})}^{- 1}} + \frac{l n (x + \sqrt{1 + x^{2})}}{2} + \frac{x \sqrt{1 + x^{2}}}{2} + c_{1}$ $H = \int (1 + x^{2}) \sqrt{x + 1} d x =_{\sqrt{x + 1} = t} \int (1 + {(t^{2} - 1)}^{2}) t (2 t d t)$ $= 2 \int t^{2} (1 + t^{4} - 2 t^{2} + 1) d t = 2 \int t^{2} (t^{4} - 2 t^{2} + 2) d t$ $= 2 \int (t^{6} - 2 t^{4} + 2 t^{2}) d t = 2 {\frac{t^{7}}{7} - \frac{2}{5} t^{5} + \frac{2}{3} t^{3}} + c_{2}$ $= \frac{2}{7} {(\sqrt{x + 1})}^{7} - \frac{4}{5} {(\sqrt{x + 1})}^{5} + \frac{4}{3} {(\sqrt{1 + x})}^{3} + c_{2}$ $I = K + H t h e v a l u e o f I i s d e t e r m i n e d .$
	Commented by maxmathsup by imad last updated on 25/Jul/18
	$error at line 5 K =(1/3)ch^3 (t) +(t/2)sh(2t)=(1/3){((e^t +e^(−t) )/2)}^3 +(t/2) +(1/2)ch(t)sh(t) =(1/(24)){ x+(√(1+x^2 ))+(x+(√(1+x^2 )))^(−1) }^3 +((ln(x+(√(1+x^2 ))))/2) +((x(√(1+x^2 )))/2) +c_1$
	$e r r o r a t l i n e 5$ $K = \frac{1}{3} {c h}^{3} (t) + \frac{t}{2} s h (2 t) = \frac{1}{3} {\frac{e^{t} + e^{- t}}{2}}^{3} + \frac{t}{2} + \frac{1}{2} c h (t) s h (t)$ $= \frac{1}{24} {x + \sqrt{1 + x^{2}} + {(x + \sqrt{1 + x^{2}})}^{- 1}}^{3} + \frac{l n (x + \sqrt{1 + x^{2})}}{2} + \frac{x \sqrt{1 + x^{2}}}{2} + c_{1}$
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