Integral 1/(sqroot(84+16x+4x^2))

[alingy1]

Could you please check this integration?

I get a different answer from WolframAlpha.

Integral of 1/(sqroot(84+16x+4x^2)) dx

Look at picture please! I apologize for the other physiology information. I have no more clean scrap paper left.

 

[pasmith]

Could you please check this integration?

I get a different answer from WolframAlpha.

Integral of 1/(sqroot(84+16x+4x^2)) dx

Look at picture please! I apologize for the other physiology information. I have no more clean scrap paper left.

So far as $$\int \frac 1{\sqrt{84 + 16x + 4x^2}}\,dx = \frac14 \int \frac{2(x+2)}{(x + 2)\sqrt{(x+ 2)^2 + 17}}\,dx$$ I agree with your calculations. You now appear to substitute $u = (x + 2)^2 + 17$, with $du = 2(x + 2)\,dx$, from which you should obtain $$\frac14 \int \frac{2(x+2)}{(x + 2)\sqrt{(x+ 2)^2 + 17}}\,dx = \frac14 \int \frac{1}{(x + 2)\sqrt{u}}\,du = \frac14 \int \frac{1}{\sqrt{u - 17}\sqrt{u}}\,du$$ rather than $$\frac14 \int \frac{1}{u\sqrt{u^2 + 17}}\,du$$ which is what you have.

It would have been better to substitute $x + 2 = \sqrt{17} \sinh \theta$ after completing the square in the denominator. The motivation for this is that $\cosh^2 \theta = 1 + \sinh^2 \theta$ and the derivative of $\sinh \theta$ is $\cosh \theta$. You therefore end up with a constant integrand.

 

[scurty]

The issue here is when you do the u substitution and plug in your values for u. You should have written

$\frac{1}{4}\displaystyle\int\frac{1}{\sqrt{u-17}\sqrt{u}}du$


I wouldn't do u-substitution for this problem. Try doing trig substitution: $x+2=\sqrt{17}\tan{\theta}$

 

[alingy1]

Hi, I have never been introduced to those wild "sinh" functions! Let me get back to you after a few minutes of sweat.

 

[alingy1]

Oh. I understand. Then, I'm in a bit of a sticky situation... Since, I cannot use sinh cosh etc.., I'm stuck with a trig sub.

 

[alingy1]

Hmm. At which step should I do the trig sub?

 

[scurty]

Hmm. At which step should I do the trig sub?

The step after you complete the square in the denominator. If you haven't learned hyberbolic trig functions, then use my suggestion. I have a feeling pasmith's substitution is a bit cleaner but there is nothing incorrect about using tangent.

 

[alingy1]

Dear Sirs/Madams,
Here is my second attempt. But, I graphed it and it does not give the same answer as the hyperbolic function that wolfram alpha gives me.

 

[scurty]

That looks good to me. What did Wolfram give as an answer? Likely, you could manipulate that answer into the answer you gave.

 

[alingy1]

ArcSinh[(2 + x)/Sqrt[17]]/2

 

[alingy1]

The functions do not match!

 

[scurty]

ArcSinh[(2 + x)/Sqrt[17]]/2

They are equivalent to each other. Plug their difference into WolframAlpha to see:

ArcSinh[(2 + x)/Sqrt[17]]/2 - 1/2log(sqrt{(x+2)^2+17}/sqrt(17)+(x+2)/sqrt(17))

A good exercise for you would be to use the definition of arcsinh to write that form into your form.

As a starting point: $\sinh^{-1}{x}=\ln{(x+\sqrt{1+x^2})}$

 

[Ray Vickson]

Could you please check this integration?

I get a different answer from WolframAlpha.

Integral of 1/(sqroot(84+16x+4x^2)) dx

Look at picture please! I apologize for the other physiology information. I have no more clean scrap paper left.

If you set $y = x+2$ you have an integral of the form
$$\int \frac{dy}{\sqrt{y^2 + a^2}} \: (a > 0)$$
(times a constant), and if you further put $y = \sqrt{a}\: u$ you end up having to do the integral
$$\int \frac{du}{\sqrt{u^2+1}}$$
You might as well do this integral once-and-for-all, then use it in your current problem.