0
$\begingroup$

This question already has an answer here:

I am learning homogeneous differential equations, and I tried to answer one of the questions, but I am stuck at integrating the equation above. Are there any easier ways than making the denominator: $$(x-\frac{x}{2})^2+\frac{7}{4}$$?

$\endgroup$

marked as duplicate by Jyrki Lahtonen, RRL, John Omielan, Misha Lavrov, Paul Frost Apr 26 at 21:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

0
$\begingroup$

Hint. One may just observe that, by the chain rule, $$ \left(\frac1b\arctan \frac{x-a}{b} \right)'_x=\frac1{(x-a)^2+b^2}. $$

$\endgroup$
0
$\begingroup$

Use that $$x^2-x+2=\left(x-\frac{1}{2}\right)^2+\frac{7}{8}$$ and substitute $$t=x-\frac{1}{2}$$

$\endgroup$

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