How do you evaluate $\int$ $int$ $\sqrt{x^{2} + 14 x}$ $sqrt(x^2 + 14x)$ dx?

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How do you evaluate $\int$ $int$ $\sqrt{x^{2} + 14 x}$ $sqrt(x^2 + 14x)$ dx?

Evaluate the integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.)

$\int$ $int$ $\sqrt{x^{2} + 14 x}$ $sqrt(x^2 + 14x)$ dx?

1 Answer

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Answer 1 · 2017-06-17T12:55:34

The answer is $= \frac{1}{2} (x + 7) \sqrt{x^{2} + 14 x} - \frac{49}{2} ln (∣ ∣ ∣ \frac{1}{7} (\sqrt{x^{2} + 14 x} + (x + 7)) ∣ ∣ ∣) + C$ $=1/2(x+7)sqrt(x^2+14x)-49/2ln(|1/7(sqrt(x^2+14x)+(x+7))|)+C$

Explanation:

We need

${cosh}^{2} θ - {sinh}^{2} θ = 1$ $cosh^2theta-sinh^2theta=1$

$cosh 2 θ = 2 {sinh}^{2} θ + 1$ $cosh 2theta=2sinh^2theta+1$

${sinh}^{2} θ = \frac{1}{2} (cosh 2 θ - 1)$ $sinh^2theta=1/2(cosh 2theta-1)$

$sinh 2 θ = 2 sinh θ cosh θ$ $sinh2theta=2sinhthetacoshtheta$

$a r c cosh x = ln (\sqrt{x^{2} - 1} + x)$ $arc cosh x=ln(sqrt(x^2-1)+x)$

We perform this integral by substitution but first we do some simplification

$x^{2} + 14 x = x^{2} + 14 x + 49 - 49 = {(x + 7)}^{2} - 49$ $x^2+14x=x^2+14x+49-49=(x+7)^2-49$

The substitution is

$x + 7 = 7 cosh θ$ $x+7=7coshtheta$

$d x = 7 sinh θ d θ$ $dx=7sinhtheta d theta$

$\sqrt{{(x + 7)}^{2} - 49} = \sqrt{49 {cosh}^{2} θ - 49}$ $sqrt((x+7)^2-49)=sqrt(49cosh^2theta-49)$

$= 7 \sqrt{{cosh}^{2} θ - 1}$ $=7sqrt(cosh^2theta-1)$

$= 7 sinh θ$ $=7sinh theta$

${sinh}^{2} θ = {cosh}^{2} θ - 1 = {(\frac{x + 7}{7})}^{2} - 1$ $Sinh^2 theta=cosh^2theta-1=((x+7)/7)^2-1$

$= \frac{x^{2} + 14 x + 49 - 49}{49}$ $=(x^2+14x+49-49)/49$

$= \frac{x^{2} + 14 x}{49}$ $=(x^2+14x)/49$

$sinh θ = \frac{1}{7} \sqrt{x^{2} + 14 x}$ $sinh theta=1/7sqrt(x^2+14x)$

Therefore,

$\int \sqrt{x^{2} + 14 x} d x = \int 7 sinh θ \cdot 7 sinh θ d θ$ $intsqrt(x^2+14x)dx=int7sinhtheta*7sinh theta d theta$

$= 49 \int {sinh}^{2} θ d θ$ $=49int sinh^2 theta d theta$

$= \frac{49}{2} \int (cosh 2 θ - 1) d θ$ $=49/2int(cosh 2theta-1) d theta$

$= \frac{49}{2} (\frac{sinh 2 θ}{2} - θ)$ $=49/2((sinh 2theta)/2-theta)$

$= \frac{49}{2} (sinh θ cosh θ - a r c cosh (\frac{x + 7}{7}))$ $=49/2(sinhthetacoshtheta-arc cosh((x+7)/7))$

$= (\frac{49}{2} \cdot \frac{1}{7} \cdot \sqrt{x^{2} + 14 x} \cdot \frac{x + 7}{7}) - \frac{49}{2} ⎛ ⎝ ln ⎛ ⎝ \sqrt{\frac{{(x + 7)}^{2}}{49} - 1} ⎞ ⎠ + \frac{x + 7}{7} ⎞ ⎠ + C$ $=(49/2*1/7*sqrt(x^2+14x)*(x+7)/7)-49/2(ln(sqrt((x+7)^2/49-1))+(x+7)/7)+C$

$= \frac{1}{2} (x + 7) \sqrt{x^{2} + 14 x} - \frac{49}{2} ln (∣ ∣ ∣ \frac{1}{7} (\sqrt{x^{2} + 14 x} + (x + 7)) ∣ ∣ ∣) + C$ $=1/2(x+7)sqrt(x^2+14x)-49/2ln(|1/7(sqrt(x^2+14x)+(x+7))|)+C$

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How do you evaluate $\int$ $int$ $\sqrt{x^{2} + 14 x}$ $sqrt(x^2 + 14x)$ dx?

Evaluate the integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.)

$\int$ $int$ $\sqrt{x^{2} + 14 x}$ $sqrt(x^2 + 14x)$ dx?

1 Answer

Explanation:

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1 Answer

Explanation:

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