Evaluate the integral or show that it is divergent ( 1 to infinity, tan-1 x/x^2 dx

[real mckoy]

I_1^INFINITY tan-1 (x)/x^2 dx I have correctly set up the limit t--> infinity for I_1^t and integrated using integration by parts. u=tan-1 x du= 1/1+x^2 dv= x^-2 dx v= -1/x

I am lost on the step: I_NTEGRAL (1/X) (dx/1+x^2)= I_NTEGRAL [(1/X) - ( X/X^2+1) dx


problem #50 ch8 review, stewart single variable calculus 6E

Thanks for the help!

 

[soroban]

Hello, real mckoy!

\(\displaystyle \displaystyle I \;=\;\int^{\infty}_1 \frac{\tan^{-1}x}{x^2}\,dx\)

By parts: .\(\displaystyle \begin{Bmatrix}u &=& \tan^{\text{-}1}x && dv &=& x^{-2}\,dx \\ du &=& \dfrac{dx}{1+x^2} && v &=& -x^{\text{-}1}\end{Bmatrix}\)

\(\displaystyle \displaystyle I \;=\;-x^{\text{-}1}\tan^{\text{-}1}x - \int^{\infty}_1\!\! \left(-x^{\text{-}1}\right)\left(\dfrac{dx}{1+x^2}\right) \;=\; -\frac{\tan^{\text{-}1}x}{x} + \int^{\infty}_1\!\! \frac{dx}{x(1+x^2)} \)


Partial Fractions:

. . \(\displaystyle \displaystyle I \;=\;-\frac{\tan^{\text{-}1}x}{x} + \int^{\infty}_1\!\!\left(\frac{1}{x} - \frac{x}{1+x^2}\right)\,dx \)

. . \(\displaystyle \displaystyle I \;=\;-\frac{\tan^{\text{-}1}x}{x} + \ln |x| - \tfrac{1}{2}\ln\left|1+x^2\right|\,\bigg]^{\infty}_1\)


I'll let you evaulate it . . .

 

[real mckoy]

soroban,

Thank you.