Recent content by Denisse

\begin{array}{ccc} 2 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 3 \end{array}

I tried with x=tan u and it seems like the integral is divergent because of this: ##\int _{ 0 }^{ \infty }{ \frac { 1 }{ x\left( { x }^{ 2 }+1 \right) } dx } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \cos { u } }{ \sin { u } } du } ##

How can I get the improper integral ## \int_0^{\infty}\frac{1}{x(1+x^2)}\,dx ## First thing I tried was separating the integral like this ## \int_0^{1}\frac{1}{x(1+x^2)}\,dx + \int_1^{\infty}\frac{1}{x(1+x^2)}\,dx## And then I tried with partial fractions but it didn't work

Is it posible that the product of two different divergent series be a convergent serie?

Perfect squares? How? I don't see it

Could you help me to find the general term of the sequence: ## 1 , \frac{5}{3} , 1 , \frac{15}{17} , 1 , \frac{37}{35} , 1 , \frac{63}{65} ,... ## Thank you!

Homework Statement How can I integrate this? I already tried substitution u=x-1 and partial fractions. ∫[dx]/[(1-X^2)√((x^2)-3x+2)] Homework Equations The Attempt at a Solution