# Finding the indefinite integral $\int\frac{\sqrt{x+8}}{\sqrt{x-3}-\sqrt{x+3}}dx$

This is a homework question, I tried many subtitutions but nothing worked for me...

$$\int\frac{\sqrt{x+8}}{\sqrt{x-3}-\sqrt{x+3}}dx$$

Any clue will help.

Thanks.

• Rationalize the denominator. Mar 31, 2013 at 11:57
• Try getting rid of the square roots on the bottom by multiplying the denominator (and numerator) by something suitable Mar 31, 2013 at 11:58

Hint::

$$\int\frac{\sqrt{x+8}}{\sqrt{x-3}-\sqrt{x+3}}dx$$ $$=\int\frac{\sqrt{x+8}(\sqrt{x-3}+\sqrt{x+3})}{(\sqrt{x-3}-\sqrt{x+3})(\sqrt{x-3}+\sqrt{x+3})}dx$$ $$=\int\frac{\sqrt{x+8}\sqrt{x-3}}{6}dx+\int\frac{\sqrt{x+8}\sqrt{x+3}}{6}dx$$ $$=\int\frac{\sqrt{x^2+5x-24}}{6}dx+\int\frac{\sqrt{x^2+5x+24}}{6}dx$$ $$=\frac{I_1+I_2}{6}$$ $$I_1 = \int\sqrt{x^2+5x-24}$$ $$I_1= \int\sqrt{\left(x+\frac{5}{2}\right)^2-\frac{121}{4}}dx$$ Substitute $u = \left(x+\frac{5}{2}\right)dx$ $$I_1= \int\sqrt{u^2-\frac{121}{4}}du$$ Substitute $\frac{11\sec\theta}{2} = u;du = \frac{11\sec\theta\tan \theta}{2}d\theta$ $$I_1= \frac{121}{4}\int\tan ^2\theta \sec \theta d\theta$$ $$I_1= \frac{121}{4}\left[\int\sec^3 \theta d\theta - \int \sec \theta d\theta\right]$$ Using reduction formula $$\int{sec^m \theta} = \frac{\sin \theta \sec ^{m-1} \theta}{m-1} + \frac{m-2}{m-1}\int\sec^{m-2}\theta d\theta$$

Substitute and Solve

• Whats going on with this first equality? It should be $$\int \frac{\sqrt{x+8}}{\sqrt{x-3}-\sqrt{x+3}}dx = \int\frac{\sqrt{8+x}(\sqrt{x-3}+\sqrt{x+3})}{(\sqrt{x-3}-\sqrt{x+3})(\sqrt{x-3} +\sqrt{x+3})}dx$$ Mar 31, 2013 at 16:08
• @stochasm: Thanks, Typographical Error, corrected. Mar 31, 2013 at 16:15

As sugested by other users: \begin{align} \int\frac{\sqrt{x+8}}{\sqrt{x-3}-\sqrt{x+3}}dx&=\int\frac{\sqrt{x+8}(\sqrt {x-3}+\sqrt{x+3})}{(\sqrt{x-3}-\sqrt{x+3})(\sqrt{x-3}+\sqrt{x+3})}dx\\ &=\int\frac{\sqrt{x^2+5x-24}+\sqrt{x^2+11x+24}}{(x-3)-(x+3)}dx \end{align}

Now let $t=\sqrt{x^2+5x-24}-x$.

Since

\begin{align} t=\sqrt{x^2+5x-24}-x &\Longrightarrow t+x=\sqrt{x^2+5x-24}\\ &\Longrightarrow x^2+2xt+t^2=x^2+5x-24\\ &\Longrightarrow 5x-2xt= t^2+24\\ &\Longrightarrow x(5-2t)=t^2+24\\ &\Longrightarrow x=\frac{t^2+24}{5-2t}\\ &\Longrightarrow dx=\frac{2t(5-2t)-(t^2+24)\cdot (-2)}{(5-2t)^2}dt\\ &\Longrightarrow dx=\frac{10t-4t^2+2t^2+48}{(2t-5)^2}dt\\ &\Longrightarrow dx=\frac{-t^2+5t+24}{2(2t-5)^2}dt \end{align}

By sucessive modus ponens you get

$$\sqrt{x^2+5x-24}=x+t \space \wedge \space x=\frac{t^2+24}{5-2t} \space \wedge \space dx=\frac{-t^2+5t+24}{2(2t-5)^2}dt$$

Therefore

\begin{align} \int \underbrace{\sqrt{x^2+5x-24}}_{\displaystyle x+t}dx&=\int \underbrace{\left(\frac{t^2+24}{5-2t}+t\right)}_{\displaystyle x+t}\cdot\underbrace{\frac{-t^2+5t+24}{2(2t-5)^2}dt}_{\displaystyle dx} \end{align}

From here onwards it's a different method of integration, which you should be familiar with.

I would like to ask other users to proofread my calculations. Thanks.

Hint: If you multiply an divide by $\sqrt{x-3}+\sqrt{x+3}$, the result will be this, getting rid of the fraction and obtainin just two integrals of a single square root. That may be easier to solve: $$\int\frac{\sqrt{x^2+5x-24}+\sqrt{x^2+11x+24}}{-6}=\int\frac{\sqrt{(x-3)(x+8)}+\sqrt{(x+3)(x+8)}}{-6}=$$ $$=\frac{-1}{6}\int\sqrt{(x-3)(x+8)}-\frac{1}{6}\int\sqrt{(x+3)(x+8)}$$

• I multiplied by what I said. $x=0$ is not part of the domain of that function becuase there's a $\sqrt{(x-3)}$ Mar 31, 2013 at 12:07
• I agree with @Git Gud
– Amr
Mar 31, 2013 at 12:08
• @GitGud I made mistakes... you're right. Mar 31, 2013 at 13:24
• Thank you all - helped me alot.
– dave
Mar 31, 2013 at 23:22