Evaluate the Improper Integral or State That It Is Divergent$$\int _ { 0 } ^ { \infty } \frac { 36 \left( 1 + \tan ^ { - 1 } x \right) } { 1 + x ^ { 2 } } d x$$

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The Answer of Evaluate the Improper Integral or State That It Is Divergent$$\int _ { 0 } ^ { \infty } \frac { 36 \left( 1 + \tan ^ { - 1 } x \right) } { 1 + x ^ { 2 } } d x$$

Evaluate the Improper Integral or State That It Is Divergent $\int _ { 0 } ^ { \infty } \frac { 36 \left( 1 + \tan ^ { - 1 } x \right) } { 1 + x ^ { 2 } } d x$

Definitions:

Evaluate the Improper Integral or State That It Is Divergent ∫0∞36(1+tan⁡−1x)1+x2dx\int _ { 0 } ^ { \infty } \frac { 36 \left( 1 + \tan ^ { - 1 } x \right) } { 1 + x ^ { 2 } } d x∫0∞​1+x236(1+tan−1x)​dx

Definitions:

Evaluate the Improper Integral or State That It Is Divergent $\int _ { 0 } ^ { \infty } \frac { 36 \left( 1 + \tan ^ { - 1 } x \right) } { 1 + x ^ { 2 } } d x$