Integral containing something over the root of x²-a²

**Ulysses** · June 30th 2010, 02:10 PM

Hi there, I'm facing this problem which I couldn't solve.

The statement says:
Evaluate the next integrals, expressing it previously in the forms that contains $\sqrt[ ]{a^2+u^2}$ , $\sqrt[ ]{a^2-u^2}$ and $\sqrt[ ]{u^2-a^2}$ , and then solve it using the integral table;

The exercise which I couldn't solve is:

$\displaystyle\int_{}^{}\displaystyle\frac{x+3}{\sq rt[ ]{x^2+2x}}dx$

I started by completing the square:

$x^2+2x=(x+1)^2-1$

Then

$u^2=(x+1)^2$
$u=(x+1)\Rightarrow{x+3=u+2}$
$du=dx$

$\displaystyle\int_{}^{}\displaystyle\frac{x+3}{\sq rt[ ]{x^2+2x}}dx=\displaystyle\int_{}^{}\displaystyle\f rac{u+2}{\sqrt[ ]{u^2-1}}du$

From here I've tried to solve it by parts
$t=u+2$
$dt=du$

$dv=\displaystyle\frac{1}{\sqrt[ ]{u^2-1}}du$
$v=\ln|u+\sqrt[ ]{u^2-1}|$

$\displaystyle\int_{}^{}\displaystyle\frac{u+2}{\sq rt[ ]{u^2-1}}du=(u+2)\ln|u+\sqrt[ ]{u^2-1}|-\displaystyle\int_{}^{}\ln|u+\sqrt[ ]{u^2-1}|du$

If I try again by parts with $-\displaystyle\int_{}^{}\ln|u+\sqrt[ ]{u^2-1}|du$ it gets more complicated.

So, what do you say?

Bye there.

**Plato** · June 30th 2010, 02:23 PM

Have tried WolframAlpha?

**pickslides** · June 30th 2010, 02:35 PM

Read this Trigonometric substitutions

**Ulysses** · June 30th 2010, 02:43 PM

Ok, I think I have solved it, here is the solution I have arrived to:

$\displaystyle\int_{}^{}\displaystyle\frac{x+3}{\sq rt[ ]{x^2+2x}}dx=\sqrt[ ]{(x+1)^2-1}+2\ln|x+1+\sqrt[ ]{(x+1)^2-1}|+C$

**TheCoffeeMachine** · June 30th 2010, 04:14 PM

Originally Posted by Ulysses

Ok, I think I have solved it, here is the solution I have arrived to:

$\displaystyle\int_{}^{}\displaystyle\frac{x+3}{\sq rt[ ]{x^2+2x}}dx=\sqrt[ ]{(x+1)^2-1}+2\ln|x+1+\sqrt[ ]{(x+1)^2-1}|+C$

That's correct!

Here is how I went about it: $\displaystyle \int\frac{x+3}{\sqrt{(x^2+2x)}}\;{dx} = \int\frac{x+3}{\sqrt{(x+1)^2-1}}\;{dx}$ . Let $x+1 = \cosh{\theta}$ , then $\dfrac{dx}{d\theta} = \sinh{\theta} \Rightarrow {dx} = \sinh{\theta}\;{d\theta}$ . Also $x+1 = \cosh{\theta} \Leftrightarrow x+3 = \cosh{\theta}+2$ . Putting this together we have $\displaystyle \int\frac{x+3}{\sqrt{(x^2+2x)}}\;{dx} = \int\frac{(\cosh{\theta}+2)(\sinh{\theta})}{\sqrt{ {\cosh^2{\theta}}-1}}\;{d\theta}$ $\displaystyle = \int\frac{(\cosh{\theta}+2)(\sinh{\theta})}{\sqrt{ \sinh^2{\theta}}}\;{d\theta} = \int\frac{(\cosh{\theta}+2)(\sinh{\theta})}{(\sinh {\theta})}\;{d\theta}$ $\displaystyle = \int\cosh{\theta}+2\;{d\theta} = \sinh{\theta}+2{\theta}+k,$ where $\theta = \text{arccosh}(x+1)$ of course. Using the little beautiful identity $\sinh\left\{\text{arccosh}\varphi\right\} = \sqrt{\varphi-1}\sqrt{\varphi+1}$ this reduces to $\sqrt{x}\sqrt{x+2}+2\text{arccosh}(x+1)+k.$

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