find ∫ (x^2 −1)(√(x^2 +1))dx


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Question Number 85166 by mathmax by abdo last updated on 19/Mar/20

$f i n d \int (x^{2} - 1) \sqrt{x^{2} + 1} d x$
Commented by john santu last updated on 21/Mar/20
$let K = ∫ x^2 (√(x^2 +1)) dx K = (1/2)∫ x { 2x(√(x^2 +1)) } dx K= (1/2) ∫ x { (√(x^2 +1)) d(x^2 +1) } = (1/2){ (2/3) x (x^2 +1)^(3/2) −∫ (x^2 +1)^(3/2) dx} = (1/3)x(x^2 +1)^(3/2) −(1/2) ∫ sec^5 u du [ x = tan u ] H= ∫ sec^5 u du = ∫ sec^3 u d(tan u) H = sec^3 u tan u −∫ tan u (3sec^3 u tan u )du H= sec^3 u tan u −3∫(sec^5 u−sec^3 u)du 4H = sec^3 u tan u +3∫ sec^3 u du H = (1/4)x (x^2 +1)^(3/2) +(3/4)sec^3 u du let J = ∫ (√(x^2 +1)) dx , [ x=tan t ] J = ∫ sec^3 t dt = ∫ sec t d(tan t) J = sec t tan t − ∫ tan^2 t sec t dt J = sec t tan t − ∫ (sec^3 t−sec t) dt 2J = sec t tan t +ln ∣ x+(√(x^2 +1)) ∣ J= (1/2)sec t tan t + (1/2)ln ∣x+(√(x^2 +1)) ∣ ∴ H = (1/4)x(x^2 +1)^(3/2) +(7/8)x(x^2 +1)^(3/2) + (7/8)ln ∣x+(√(x^2 +1)) ∣ +c K= (1/3)x(x^2 +1)^(3/2) −(1/5)H ∴ K =(1/3)x(x^2 +1)^(3/2) −(9/(40))x(x^2 +1)^(3/2) − (7/(40)) ln ∣x+(√(x^2 +1)) ∣ +c K = ((13)/(120))x(x^2 +1)^(3/2) −(7/(40))ln ∣x+(√(x^2 +1)) ∣ + c$
$let K = \int x^{2} \sqrt{x^{2} + 1} dx$ $K = \frac{1}{2} \int x {2 x \sqrt{x^{2} + 1}} dx$ $K = \frac{1}{2} \int x {\sqrt{x^{2} + 1} d (x^{2} + 1)}$ $= \frac{1}{2} {\frac{2}{3} x {(x^{2} + 1)}^{\frac{3}{2}} - \int {(x^{2} + 1)}^{\frac{3}{2}} dx}$ $= \frac{1}{3} x {(x^{2} + 1)}^{\frac{3}{2}} - \frac{1}{2} \int \sec^{5} u du$ $[x = \tan u]$ $H = \int \sec^{5} u du = \int \sec^{3} u d (\tan u)$ $H = \sec^{3} u \tan u - \int \tan u (3 \sec^{3} u \tan u) du$ $H = \sec^{3} u \tan u - 3 \int (\sec^{5} u - \sec^{3} u) du$ $4 H = \sec^{3} u \tan u + 3 \int \sec^{3} u du$ $H = \frac{1}{4} x {(x^{2} + 1)}^{\frac{3}{2}} + \frac{3}{4} \sec^{3} u du$ $let J = \int \sqrt{x^{2} + 1} dx, [x = \tan t]$ $J = \int \sec^{3} t dt = \int \sec t d (\tan t)$ $J = \sec t \tan t - \int \tan^{2} t \sec t dt$ $J = \sec t \tan t - \int (\sec^{3} t - \sec t) dt$ $2 J = \sec t \tan t + \ln ∣ x + \sqrt{x^{2} + 1} ∣$ $J = \frac{1}{2} \sec t \tan t + \frac{1}{2} \ln ∣ x + \sqrt{x^{2} + 1} ∣$ $∴ H = \frac{1}{4} x {(x^{2} + 1)}^{\frac{3}{2}} + \frac{7}{8} x {(x^{2} + 1)}^{\frac{3}{2}} +$ $\frac{7}{8} \ln ∣ x + \sqrt{x^{2} + 1} ∣ + c$ $K = \frac{1}{3} x {(x^{2} + 1)}^{\frac{3}{2}} - \frac{1}{5} H$ $∴ K = \frac{1}{3} x {(x^{2} + 1)}^{\frac{3}{2}} - \frac{9}{40} x {(x^{2} + 1)}^{\frac{3}{2}} -$ $\frac{7}{40} \ln ∣ x + \sqrt{x^{2} + 1} ∣ + c$ $K = \frac{13}{120} x {(x^{2} + 1)}^{\frac{3}{2}} - \frac{7}{40} \ln ∣ x + \sqrt{x^{2} + 1} ∣ + c$
Commented by mathmax by abdo last updated on 21/Mar/20

$t h a n k y o u s i r j o h n$
Commented by mathmax by abdo last updated on 21/Mar/20

$A = \int (x^{2} - 1) \sqrt{x^{2} + 1} d x f o r t h i s k i n d o f i n t e g r a l s w e d o t h e$ $c h a n g e m e n t x = s h (t) \Rightarrow A = \int ({s h}^{2} t - 1) c h (t) c h (t) d t$ $A = \int ({s h}^{2} t - 1) {c h}^{2} t d t = \int (\frac{c h (2 t) - 1}{2} - 1) (\frac{c h (2 t) + 1}{2}) d t$ $= \frac{1}{4} \int (c h (2 t) - 3) (c h (2 t) + 1) d t$ $= \frac{1}{4} \int ({c h}^{2} (2 t) + c h (2 t) - 3 c h (2 t) - 3) d t$ $= \frac{1}{4} \int ({c h}^{2} (2 t) - 2 c h (2 t) - 3) d t$ $= \frac{1}{8} \int (1 + c h (4 t)) d t - \frac{1}{2} \int c h (2 t) - \frac{3}{4} t + C$ $= \frac{1}{8} t + \frac{1}{32} s h (4 t) - \frac{1}{4} s h (2 t) - \frac{3}{4} t + C$ $= - \frac{5}{8} t + \frac{1}{16} s h (2 t) c h (2 t) - \frac{1}{2} s h (2 t) c h (2 t) + C$ $= - \frac{5}{8} t + \frac{1}{8} s h (t) c h (t) (1 + 2 {s h}^{2} t) - s h (t) {c h}^{2} t + C$ $= - \frac{5}{8} l n (x + \sqrt{1 + x^{2}}) + \frac{1}{8} x \sqrt{1 + x^{2}} (1 + 2 x^{2}) - x (1 + x^{2}) + C$
	Answered by MJS last updated on 20/Mar/20

	$\int (x^{2} - 1) \sqrt{x^{2} + 1} d x =$ $[t = x + \sqrt{x^{2} + 1} \to d x = \frac{\sqrt{x^{2} + 1}}{x + \sqrt{x^{2} + 1}} d t]$ $= \int (\frac{t^{3}}{16} - \frac{t}{4} - \frac{5}{8 t} - \frac{1}{4 t^{3}} + \frac{1}{16 t^{5}}) d t =$ $= \frac{t^{4}}{64} - \frac{t^{2}}{8} - \frac{5}{8} \ln t + \frac{1}{8 t^{2}} - \frac{1}{64 t^{4}} =$ $= \frac{t^{8} - 8 t^{6} + 8 t^{2} - 1}{64 t^{4}} - \frac{5}{8} \ln t =$ $= \frac{x (2 x^{2} - 3) \sqrt{x^{2} + 1}}{8} - \frac{5}{8} \ln (x + \sqrt{x^{2} + 1}) + C$
	Commented by mathmax by abdo last updated on 20/Mar/20

	$t h a n k y o u s i r m j s$
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