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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

$\int \frac{d x}{x^{2 / 3}\left(1+x^{2 / 3}\right)}=$

$\int \frac{d x}{(x-3)^{\frac{4}{5}}(x+1)^{\frac{6}{5}}} = $

Find $\int \frac{(3 \sin \phi-2) \cos \phi}{5-\cos ^{2} \phi-4 \sin \phi} d \phi$

$\int {\frac{t}{e^{3t^2}}} \, dt = $

If $f_n(x) = \log \log \log \ldots \log x$ (where $\log$ is repeated $n$ times), then $\int (x f_1(x) f_2(x) \ldots f_n(x))^{-1} dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module