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Could someone list out the steps, I managed to find the equation in terms of x. But im having trouble integrating sqrt (x^2+4x)

 May 3, 2017
 #1
avatar+100005 
+1

This has already been answered here

 

https://web2.0calc.com/questions/one-more

 May 3, 2017
 #2
avatar+21977 
+2

Need help with this continuous fraction integral.

integral \limits_{x=0}^1  \frac{x\pm\sqrt{x^2+4x} }{2}  \ dx

 

Continuous Fraction.

\(\begin{array}{|rcll|} \hline \begin{equation*} sum=x+\cfrac{x}{x+\cfrac{x}{x+\cfrac{x}{x+\cdots}}} \end{equation*}\\\\ sum &=& x + \frac{x}{sum} \\ sum - \frac{x}{sum} &=& x \\ \frac{sum^2-x}{sum} &=& x \\ sum^2-x &=& x\cdot sum \\ sum^2-x\cdot sum -x &=& 0 \\ sum &=& \frac{x\pm\sqrt{x^2-4\cdot(-x)} }{2} \\ \mathbf{sum} & \mathbf{=} & \mathbf{ \frac{x\pm\sqrt{x^2+4x} }{2} } \\ \hline \end{array} \)

 

\(\small{ \begin{array}{llcl} \int \limits_{x=0}^{1} { x+\cfrac{x}{x+\cfrac{x}{x+\cfrac{x}{x+\cdots}}} \ dx} \\\\ = \int \limits_{x=0}^{1} { sum \ dx} \\ = \int \limits_{x=0}^{1} { \mathbf{ \frac{x\pm\sqrt{x^2+4x} }{2} } \ dx} \\ = \frac12 \int \limits_{x=0}^{1} {x \ dx } \pm \frac12 \int \limits_{x=0}^{1} { \sqrt{x^2+4x} \ dx } \\ = \frac12 \left[\frac{x^2}{2}\right]_{x=0}^{1} \pm \frac12 \int \limits_{x=0}^{1}{ \sqrt{(x+2)^2-4} \ dx } \\ = \frac14 \pm \frac12 \int \limits_{x=0}^{1}{ \sqrt{(x+2)^2-4} \ dx } \\ & \text{substitute:}\\ & \boxed{~ u=x+2\\ du = dx ~}\\ & \text{new limits:}\\ & \boxed{~ x=0: \qquad u=0+2 \Rightarrow u = 2 \\ x=1: \qquad u=1+2 \Rightarrow u = 3 ~}\\ = \frac14 \pm \frac12 \int \limits_{u=2}^{3} { \sqrt{u^2-4} \ du } \\ & \text{substitute:}\\ & \boxed{~ u=2 \cosh(z) \qquad z = \text{arcosh} \left(\frac{u}{2}\right)\\ du = 2 \sinh(z)\ dz ~}\\ & \text{new limits:}\\ & \boxed{~ u=2: \qquad z=\text{arcosh} \left(\frac{2}{2}\right) \\ \Rightarrow z = \text{arcosh}(1) = 0 \\ u=3: \qquad z=\text{arcosh} \left(\frac{3}{2}\right) \\ \Rightarrow z = \text{arcosh}(1.5) ~}\\ = \frac14 \pm \frac12 \int \limits_{z=0}^{\text{arcosh}(1.5)} { \sqrt{4\cosh(z)^2-4} \cdot 2 \cdot \sinh(z) \ dz } \\ = \frac14 \pm \int \limits_{z=0}^{\text{arcosh}(1.5)} { \sqrt{4\cosh(z)^2-4} \cdot \sinh(z) \ dz } \\ & \boxed{~ \cosh^2(z) - \sinh^2(z) = 1 \\ \cosh^2(z) - 1 = \sinh^2(z) \quad | \quad \cdot 4 \\ 4\cosh^2(z) - 4 = 4\sinh^2(z) \\ ~}\\ = \frac14 \pm \int \limits_{z=0}^{\text{arcosh}(1.5)} { \sqrt{4\sinh^2(z)} \cdot \sinh(z) \ dz } \\ = \frac14 \pm \int \limits_{z=0}^{\text{arcosh}(1.5)} { 2\sinh(z) \cdot \sinh(z) \ dz } \\ = \frac14 \pm \int \limits_{z=0}^{\text{arcosh}(1.5)} { 2\sinh^2(z) \ dz } \\ \end{array} } \)

\(\small{ \begin{array}{llcl} & \boxed{~ \cosh(2z) = \cosh^2(z) + \sinh^2(z) \\ \cosh(2z) = 1+\sinh^2(z) + \sinh^2(z) \\ \cosh(2z) = 1+2\sinh^2(z) \\ 2\sinh^2(z) = \cosh(2z)-1 ~}\\ = \frac14 \pm \int \limits_{z=0}^{\text{arcosh}(1.5)} { \Big( \cosh(2z)-1 \Big) \ dz } \\ = \frac14 \pm \Big( \int \limits_{z=0}^{\text{arcosh}(1.5)} {\cosh(2z) \ dz } -\int \limits_{z=0}^{\text{arcosh}(1.5)} {\ dz } \Big) \\ = \frac14 \pm \Big( \left[\frac12\cdot \sinh(2z) \right]_{z=0}^{\text{arcosh}(1.5)} -\left[ z \right]_{z=0}^{\text{arcosh}(1.5)} \Big) \\ & \boxed{~ \sinh(2z) = 2\sinh(z)\cosh(z) ~}\\ = \frac14 \pm \Big( \left[\frac12\cdot 2\sinh(z)\cosh(z) \right]_{z=0}^{\text{arcosh}(1.5)} - \text{arcosh}(1.5) \Big) \\ = \frac14 \pm \Big( \left[ \sinh(z)\cosh(z) \right]_{z=0}^{\text{arcosh}(1.5)} - \text{arcosh}(1.5) \Big) \\ = \frac14 \pm \Big[ \sinh\Big(\text{arcosh}(1.5)\Big)\cdot 1.5 - \text{arcosh}(1.5) \Big] \\ & \boxed{~ \cosh^2(x) - \sinh^2(x) = 1 \\ \sinh^2(x) = \cosh^2(x) - 1 \\ \sinh(x) = \sqrt{\cosh^2(x) - 1} \\ ~}\\ \boxed{~ \sinh\Big(\text{arcosh}(x)\Big)= \sqrt{\cosh^2(\text{arcosh}(x)) - 1} = \sqrt{x^2 - 1} \\ \sinh\Big(\text{arcosh}(1.5)\Big) = \sqrt{1.5^2 - 1} = \sqrt{1.25} = \frac{ \sqrt{5} }{2} ~}\\ = \frac14 \pm \Big[ \frac{\sqrt{5}}{2} \cdot \frac{3}{2} - \text{arcosh}(1.5) \Big] \\ = \frac14 \pm \Big[ \frac{3\sqrt{5}}{4} - \text{arcosh}(1.5) \Big] \\ = \frac14 \pm ( 1.6770509831248424 - 0.9624236501192069 ) \\ = \frac14 \pm 0.7146273330056354 \\ \end{array} }\)

 

\(\begin{array}{|rcll|} \hline = \frac14 + 0.7146273330056354 \qquad &\text{or}&\qquad = \frac14 - 0.7146273330056354 \\ = 0.9646273330056354 \qquad &\text{or}& \qquad =-0.4646273330056354 \\ \hline \end{array}\)

 

laugh

 May 4, 2017
edited by heureka  May 4, 2017

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