int {frac{{{e^{m{{tan }^{ - 1}}x}}}}{{1 + {x^ | vedclass

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Question

(C) Let $I = \int {\frac{{{e^{m{{	an }^{ - 1}}x}}}}{{1 + {x^2}}}dx} $.Substitute $t = m{{	an }^{ - 1}}x$.Then,differentiating both sides with respec

Vedclass · Accepted Answer

$\frac{1}{m}{e^{m{{	an }^{ - 1}}x}}$

Vedclass · Answer

(C) Let $I = \int {\frac{{{e^{m{{	an }^{ - 1}}x}}}}{{1 + {x^2}}}dx} $.Substitute $t = m{{	an }^{ - 1}}x$.Then,differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = \frac{m}{1 + x^2}$,which implies $\frac{dx}{1 + x^2} = \frac{dt}{m}$.Substituting these into the integral:$I = \int {e^t \cdot \frac{dt}{m}} = \frac{1}{m} \int {e^t dt}$.Integrating $e^t$ with respect to $t$ gives $e^t + C$.Therefore,$I = \frac{1}{m} e^t + C = \frac{1}{m} e^{m{{	an }^{ - 1}}x} + C$.

$\int {\frac{{{e^{m{{\tan }^{ - 1}}x}}}}{{1 + {x^2}}}dx} $ is equal to

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