How do you determine if the improper integral converges or diverges $\int [\frac{x arctan x}{{(1 + x^{2})}^{2}}] d x$ $int [ (x arctan x) / (1+x^2)^2 ] dx$ from negative 0 to infinity?

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How do you determine if the improper integral converges or diverges $\int [\frac{x arctan x}{{(1 + x^{2})}^{2}}] d x$ $int [ (x arctan x) / (1+x^2)^2 ] dx$ from negative 0 to infinity?

Since "negative $0$ $0$ " is $0$ $0$ , we deal with the problem as $\int_{0}^{\infty} \frac{x arctan x}{{(1 + x^{2})}^{2}} d x$ $int_0^oo(xarctanx)/(1+x^2)^2dx$

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Answer 1 · 2016-08-15T18:26:29

Given Improper Integral converges to $\frac{π}{8}$ $pi/8$ .

Explanation:

Let $I = \int_{0}^{\infty} \frac{x arctan x}{{(1 + x^{2})}^{2}} d x$ $I=int_0^oo (xarctanx)/(1+x^2)^2dx$ .

We subst. $x = tan t \Rightarrow arctan x = t, and, d x = {sec}^{2} t d t$ $x=tant rArr arctanx=t, and, dx=sec^2tdt$ .

Also, $\frac{x arctan x}{{(1 + x^{2})}^{2}} = \frac{(t) (tan t)}{{(1 + {tan}^{2} t)}^{2}} = \frac{t \cdot tan t}{{sec}^{4} t}$ $(xarctanx)/(1+x^2)^2=((t)(tant))/(1+tan^2t)^2=(t*tant)/sec^4t$

Further, $x = 0 \Rightarrow tan t = 0 \Rightarrow t = 0, &, x \to \infty, t \to \frac{π}{2}$ $x=0rArrtant=0rArrt=0, &, xrarroo, trarrpi/2$

$:. I=int_0^(pi/2) (t*tant)/sec^4t*sec^2tdt=int_0^(pi/2) t*sint/cost*cos^2tdt$

$=int_0^(pi/2) tsintcostdt=1/2inttsin(2t)dt$

$=1/2[t(-cos(2t)/2]_0^(pi/2)-1/2int_0^(pi/2)1*(-cos(2t)/2)dt$

$=-1/4[pi/2*cos(2*pi/2)-0]+1/4int_0^(pi/2)cos(2t)dt$

$=-1/4(-pi/2)+1/4[sin(2t)/2]_0^(pi/2)$

$=pi/8+1/8[sinpi-sin0]$

$=pi/8$ .

Hence, the given Improper Integral $I$ converges to $pi/8$ .

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How do you determine if the improper integral converges or diverges $\int [\frac{x arctan x}{{(1 + x^{2})}^{2}}] d x$ $int [ (x arctan x) / (1+x^2)^2 ] dx$ from negative 0 to infinity?

Since "negative $0$ $0$ " is $0$ $0$ , we deal with the problem as $\int_{0}^{\infty} \frac{x arctan x}{{(1 + x^{2})}^{2}} d x$ $int_0^oo(xarctanx)/(1+x^2)^2dx$

1 Answer

Explanation:

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How do you determine if the improper integral converges or diverges int [ (x arctan x) / (1+x^2)^2 ] dx∫[xarctanx(1+x2)2]dxint [ (x arctan x) / (1+x^2)^2 ] dx from negative 0 to infinity?

Since "negative 000" is 000, we deal with the problem as int_0^oo(xarctanx)/(1+x^2)^2dx∫∞0xarctanx(1+x2)2dxint_0^oo(xarctanx)/(1+x^2)^2dx

1 Answer

Explanation:

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How do you determine if the improper integral converges or diverges $\int [\frac{x arctan x}{{(1 + x^{2})}^{2}}] d x$ $int [ (x arctan x) / (1+x^2)^2 ] dx$ from negative 0 to infinity?

Since "negative $0$ $0$ " is $0$ $0$ , we deal with the problem as $\int_{0}^{\infty} \frac{x arctan x}{{(1 + x^{2})}^{2}} d x$ $int_0^oo(xarctanx)/(1+x^2)^2dx$