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I am preparing for an exam and found this integral in a previous test. Did I do it correctly?

My attempt.

$$ \int\frac{\sqrt{x^2+2x-3}}{x+1}\,dx $$ Complete the square of $x^2+2x-3$; I changed the integral to $$ \int\frac{\sqrt{(x-1)^2-4}}{x+1}\,dx $$ then set $u=x+1$ to get $$ \int\frac{\sqrt{u^2-4}}{u}\,dx $$ Using the triangle, $2\sec\theta=u$ and $du=2\sec\theta\tan\theta d\theta$ $$ \int\frac{\sqrt{(2\sec\theta)^2-2^2}}{2\sec\theta}2\sec\theta \tan\theta\, d\theta $$ This I simplified to $$ 2\int\tan^2\theta\, d\theta = 2\int\sec^2\theta-1\,d\theta =2[\tan\theta-\theta]+C $$

Back substitute $$ \theta=\tan^{-1}\frac{\sqrt{u^2-4}}{2} $$ and $$ \tan\theta=\frac{\sqrt{u^2-4}}{2} $$ Back substitute $u=x+1$ $$ \int\frac{\sqrt{x^2+2x-3}}{x+1}\,dx= \sqrt{(x-1)^2-4}-\tan^{-1}{\sqrt{(x-1)^2-4}}+C $$

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  • $\begingroup$ If you just need a verification. You can use wolfram alpha $\endgroup$ – MrYouMath May 13 '16 at 12:28
  • $\begingroup$ if the sign of $2x$ is right, you your mistaken has started in first line. $\endgroup$ – Guilherme Thompson May 13 '16 at 12:34
  • $\begingroup$ In the third step when u take u=x+1 , Why did u put u in place of (x-1)^2 as u^2?You have written (x-1)^2 instead of (x+1)^2 $\endgroup$ – Murtuza Vadharia May 13 '16 at 12:34
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I changed the integral to $\int\frac{\sqrt{(x-1)^2-4}}{x+1}dx$ then $u=x+1$

It should be $$\int\frac{\sqrt{(x\color{red}{+}1)^2-4}}{x+1}dx$$

I simplified to $2\int tan^2\theta d\theta$ = $2\int\sec^2\theta-1d\theta$ $=2[tan\theta-\theta]+C$

back substitute $\theta=tan^{-1}\frac{\sqrt{u^2-4}}{2}$and $tan\theta=\frac{\sqrt{u^2-4}}{2}$ back substitute $u=x+1$

I think you did nothing wrong here.

$\int\frac{\sqrt{x^2+2x-3}}{x+1}dx={\sqrt{(x-1)^2-4}}-tan^{-1}{\sqrt{(x-1)^2-4}}+C$

This is not correct :
$$\int\frac{\sqrt{x^2+2x-3}}{x+1}dx={\sqrt{(x\color{red}{+}1)^2-4}}-\color{red}{2}\tan^{-1}\frac{\sqrt{(x\color{red}{+}1)^2-4}}{\color{red}{2}}+C$$

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  • $\begingroup$ than you @mathlove $\endgroup$ – Gobabis May 13 '16 at 12:44
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You've made a mistake. I hope you can find it using my answer

$$\int\frac{\sqrt{x^2+2x-3}}{x+1}\space\text{d}x=\int\frac{\sqrt{(x+1)^2-4}}{x+1}\space\text{d}x=$$


Substitute $u=x+1$ and $\text{d}u=\text{d}x$:


$$\int\frac{\sqrt{u^2-4}}{u}\space\text{d}u=$$


Substitute $u=2\sec(s)$ and $\text{d}u=2\tan(s)\sec(s)\space\text{d}s$.

We get that $\sqrt{u^2-4}=\sqrt{4\sec^2(s)-4}=2\tan(s)$ and $s=\text{arcsec}\left(\frac{u}{2}\right)$:


$$2\int\tan^2(s)\space\text{d}s=2\int\left[\sec^2(s)-1\right]\space\text{d}s=2\left[\int\sec^2(s)\space\text{d}s-\int1\space\text{d}s\right]$$

Notice now, the integral of $\sec^2(s)$ is equal to $\tan(s)$ and the integral of $1$ is just $s$.

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