intlimits_1^e {left( {frac{{{{tan }^{ - 1}}x} | vedclass

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(B) Let $I = \int_1^e \left( \frac{	an^{-1}x}{x} + \frac{\ln x}{1+x^2} ight) dx$. We can split this into two integrals: $I = \int_1^e \frac{	an^{-

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$	an^{-1}e$

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(B) Let $I = \int_1^e \left( \frac{	an^{-1}x}{x} + \frac{\ln x}{1+x^2} ight) dx$. We can split this into two integrals: $I = \int_1^e \frac{	an^{-1}x}{x} dx + \int_1^e \frac{\ln x}{1+x^2} dx$. Applying integration by parts to the first integral $\int_1^e 	an^{-1}x \cdot \frac{1}{x} dx$: Let $u = 	an^{-1}x$ and $dv = \frac{1}{x} dx$. Then $du = \frac{1}{1+x^2} dx$ and $v = \ln x$. Using the formula $\int u dv = uv - \int v du$: $\int_1^e \frac{	an^{-1}x}{x} dx = [	an^{-1}x \cdot \ln x]_1^e - \int_1^e \frac{\ln x}{1+x^2} dx$. Substituting this back into the expression for $I$: $I = \left( [	an^{-1}x \cdot \ln x]_1^e - \int_1^e \frac{\ln x}{1+x^2} dx ight) + \int_1^e \frac{\ln x}{1+x^2} dx$. The integral terms cancel out: $I = [	an^{-1}x \cdot \ln x]_1^e$. Evaluating the limits: $I = (	an^{-1}e \cdot \ln e) - (	an^{-1}1 \cdot \ln 1)$. Since $\ln e = 1$ and $\ln 1 = 0$: $I = 	an^{-1}e \cdot 1 - 0 = 	an^{-1}e$.

$x$	$1$	$2$	$3$	$4$
$y$	$0.7111$	$0.7222$	$0.7333$	$0.7444$

$\int\limits_1^e {\left( {\frac{{{{\tan }^{ - 1}}x}}{x} + \frac{{\ln x}}{{1 + {x^2}}}} \right)} \,dx$ is equal to

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$x$ $1$ $2$ $3$ $4$
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