2
$\begingroup$

I'm trying to solve this irrational integral $$ \int \frac{x-4}{\sqrt{x^2-4x+5}}\, dx$$ doing the substitution

$$ x= \frac{5-t^2}{2 \cdot (2+t)}$$ according to the rule.

So the integral becomes:

$$ \int \frac{1}{2} \cdot \frac{t^2+8t+11}{t^2+4t+5}\, dt =\frac{1}{2} \int 1+\frac{4t+7}{t^2+4t+5}\, dt= \frac{1}{2}t+2 \ln (t+2)+\frac{1}{2}\frac{1}{t+2} + cost$$ with $t=-x+ \sqrt{x^2-4x+5}$

The final result according to my book is instead $\sqrt{x^2-4x+5}-2 \ln( x-2+ \sqrt{(x-2)^2+1})$

I don't understant why this difference. Can someone show me where I'm making mistakes?

$\endgroup$
4
  • $\begingroup$ First thing I would do is differentiate your answer, and differentiate the book answer, and see whether you get back the original integrand. $\endgroup$ Apr 18 at 0:33
  • $\begingroup$ In the final result of the book, you forgot the ln parenthesis. $\endgroup$ Apr 18 at 0:44
  • $\begingroup$ yes, I changed the parenthesis $\endgroup$
    – Anne
    Apr 18 at 0:46
  • $\begingroup$ Is my substitution correct in the original integral? $\endgroup$
    – Anne
    Apr 18 at 0:53
2
$\begingroup$

To check your results, here's an other approach

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

$$\sqrt{x^2-4x+5}-2\int \frac{dx}{\sqrt{(x-2)^2+1}}$$

put $$x-2=\sinh(t)=\frac{e^t-e^{-t}}{2}$$ the last integrale becomes $$\int dt=t+C$$

but $$e^{2t}-2(x-2)e^t-1=0$$ gives

$$e^t=(x-2)+\sqrt{x^2-4x+5}$$

So, the final result is $$\sqrt{x^2-4x+5}-$$ $$2\ln\Bigl((x-2)+\sqrt{x^2-4x+5}\Bigr)+C$$

$\endgroup$
7
  • $\begingroup$ why $$x-2=\sinh(t)=\frac{e^t-e^{-t}}{2}$$? $\endgroup$
    – Anne
    Apr 18 at 0:52
  • $\begingroup$ @Anne I made that substitution to take benefit from $1+\sinh^2=\cosh^2$. Your book is right. how did you get the $cos(t)$. $\endgroup$ Apr 18 at 0:53
  • $\begingroup$ great! Now I have the same result of the book. What I don't understant is where I made mistakes in the substitution... $\endgroup$
    – Anne
    Apr 18 at 1:10
  • $\begingroup$ Your denominator $t^2+4t+4$ should be $ t^2+4t+5$ $\endgroup$ Apr 18 at 1:32
  • $\begingroup$ ahh ok I didn't write well but it's $t^2+4t+5$ $\endgroup$
    – Anne
    Apr 18 at 1:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.