Jump to content

question about trig substitution

Featured Replies

I completed the square in the denominator and used the substitution [imath]\tan\theta =x+\frac{3}{2}[/imath]. This gave me

 

[math]\int \frac{x}{4x^2+12x+13}dx=-\frac{1}{4}\ln\left(\frac{2}{\sqrt{4x^2+12x+13}}\right)-\frac{3}{8}\tan^{-1}\left(x+\frac{3}{2}\right)+c[/math]

 

[math]=\frac{1}{4}\ln\left(\frac{\sqrt{4x^2+12x+13}}{2}\right)-\frac{3}{8}\tan^{-1}\left(x+\frac{3}{2}\right)+c[/math]

 

I differentiated this, and it gave me the correct original integrand. Maybe mathematica removed the power of a half or something from the log.

Archived

This topic is now archived and is closed to further replies.

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.

Configure browser push notifications

Chrome (Android)
  1. Tap the lock icon next to the address bar.
  2. Tap Permissions → Notifications.
  3. Adjust your preference.
Chrome (Desktop)
  1. Click the padlock icon in the address bar.
  2. Select Site settings.
  3. Find Notifications and adjust your preference.