int frac{e^{tan ^{-1} 2 x}}{1+4 x^2} dx = | vedclass

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Question

(A) Let $I = \int \frac{e^{	an ^{-1} 2 x}}{1+4 x^2} dx$. Substitute $u = 	an ^{-1} 2 x$. Then,$du = \frac{1}{1+(2x)^2} \cdot \frac{d}{dx}(2x) dx = \

Vedclass · Accepted Answer

$\frac{1}{2} e^{	an ^{-1} 2 x}+c$

Vedclass · Answer

(A) Let $I = \int \frac{e^{	an ^{-1} 2 x}}{1+4 x^2} dx$. Substitute $u = 	an ^{-1} 2 x$. Then,$du = \frac{1}{1+(2x)^2} \cdot \frac{d}{dx}(2x) dx = \frac{2}{1+4x^2} dx$. This implies $\frac{dx}{1+4x^2} = \frac{1}{2} du$. Substituting these into the integral,we get: $I = \int e^u \cdot \frac{1}{2} du = \frac{1}{2} \int e^u du$. $I = \frac{1}{2} e^u + c$. Substituting $u = 	an ^{-1} 2 x$ back,we get: $I = \frac{1}{2} e^{	an ^{-1} 2 x} + c$.

$\int \frac{e^{\tan ^{-1} 2 x}}{1+4 x^2} dx =$

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