Find the value of $\int_1^2\frac{x^2-1}{(x^2+1+3x)(x^2-x+1)}dx$

Question

Find the value of

$\int_1^2\frac{x^2-1}{(x^2+1+3x)(x^2-x+1)}dx$

I tried but it was failed. Can you help me!

The brute force method: separate into partial fractions. If you like complex numbers, then this becomes some logarithms. If you prefer to let irreducible quadratics stay, then a little arctrig might arise. Regardless, this approach will work. — davidlowryduda, Jun 18 '15 at 08:15

Indominus · Accepted Answer · 2015-06-18T08:28:53.827

4

Hint: $\frac{x^2-1}{(x^2+1+3x)(x^2-x+1)}=\frac{2x+3}{x^2+3x+1}-\frac{2x-1}{x^2-x+1}$.

The solution is actually very clean now.

edited Jun 18 '15 at 08:28

answered Jun 18 '15 at 08:16

Indominus

Sorry. It's $x^2-1$. Edited. – Road Human Jun 18 '15 at 08:19
1

Edited. It actually makes sense to be $x^2-1$. It makes the problem has a much cleaner solution than it would be if it were $x^2+1$. Let me know if you can finish it by yourself – Indominus Jun 18 '15 at 08:30

1 Answers1