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

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

Question

(A) Let $I = \int {{e^{{{	an }^{ - 1}}x}}} \left( {\frac{{1 + x + {x^2}}}{{1 + {x^2}}}} ight)dx$. Substitute ${	an ^{ - 1}}x = t$,which implies $x

Vedclass · Accepted Answer

$x{e^{{{	an }^{ - 1}}x}} + c$

Vedclass · Answer

(A) Let $I = \int {{e^{{{	an }^{ - 1}}x}}} \left( {\frac{{1 + x + {x^2}}}{{1 + {x^2}}}} ight)dx$. Substitute ${	an ^{ - 1}}x = t$,which implies $x = 	an t$ and $dx = \sec^2 t \, dt$. Also,$\frac{dx}{1+x^2} = dt$. Substituting these into the integral: $I = \int e^t (1 + 	an t + 	an^2 t) dt$ $I = \int e^t (1 + 	an^2 t + 	an t) dt$ Since $1 + 	an^2 t = \sec^2 t$,we have: $I = \int e^t (\sec^2 t + 	an t) dt$. Using the standard integral formula $\int e^t (f(t) + f'(t)) dt = e^t f(t) + C$,where $f(t) = 	an t$ and $f'(t) = \sec^2 t$: $I = e^t 	an t + C$. Substituting back $t = 	an^{-1} x$: $I = x e^{	an^{-1} x} + C$.

$\int {{e^{{{\tan }^{ - 1}}x}}} \left( {\frac{{1 + x + {x^2}}}{{1 + {x^2}}}} \right)dx$ is equal to

Similar Questions

Integrate the function: $e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right)$

The value of $\int e^{x}\left[\frac{1+\sin x}{1+\cos x}\right] d x$ is equal to

$\int e^{x} \sec x(1+\tan x) d x$ equals

$\int {{e^x} \left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)} \,dx = $

$\int {{e^{\sin x}}\left( {\sin x + {{\sec }^2}x} \right)} \,dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module