evaluate-the-improper-integral-18tan1x1x2-dxdisplaystyle-int1infty-frac8-tan-1-x

Question

Evaluate the improper integral ∫1∞8tan⁡−1x1+x2 dx\displaystyle \$$int_1$$^{\infty} \frac{8 \$$tan^{-1} x$$}{1 + $$x^2$$} \, dx.

Accepted Answer

3π24{\displaystyle\frac{3\$$pi^2$$}{4}}

Answer

π24{{\displaystyle\frac{\$$pi^2$$}{4}}}

Answer

3π22\displaystyle\frac{3\$$pi^2$$}{2}

Answer

The integral diverges.

Evaluate the improper integral ∫1∞8tan−1x1+x2dx\(\displaystyle\) \(\int\)_1^{\(\infty\)} \(\frac{8 \tan^{-1}\) x}{1 + x^2} \, dx.