Integral of type 'derivative over function' with a twist

BalintRigo
Messages
7
Reaction score
1

Homework Statement



Find the integral

[(2x - 3) / (x^2 - 3x - 5)^2] dx

Homework Equations



I noticed that if I differentiate the denominator I get the nominator, which would be a simple problem. The denominator, however is raised to the power 2.

Can I still somehow use the rule for integrals where the nominator is the derivative of the denominator? Or do I need to take a different approach?

Thank you
 
Physics news on Phys.org
The 'approach' you are talking about is integration by substitution, I hope. You've observed that if u=x^2-3x-5, then du=u'dx=(2x-3)dx. That's great. That turns the integral into du/u^2 in terms of u. Can you integrate that?
 
I got it, thanks an awful lot!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top