What is the integral from 2 to 1/x of arctan(t) ?

The integral from 2 to 1/x of arctan(t) is (arctan(1/x))/x-1/2 ln| 1/(x^2)+1|-2arctan(2)+1/2 ln(5)

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integral from 2 to 1/x of arctan(t)

\int_{2}^{\frac{1}{x}} arctan (t) d t

\frac{arctan ( \frac{1}{x} )}{x} - \frac{1}{2} ln \frac{1}{x ^{2}} + 1 - 2 arctan (2) + \frac{1}{2} ln (5)

Solution steps

\int_{2}^{\frac{1}{x}} arctan (t) d t

Apply Integration By Parts

: [t arctan (t) - \int \frac{t}{t ^{2} + 1} d t]_{2}^{\frac{1}{x}}

= [t arctan (t) - \int \frac{t}{t ^{2} + 1} d t]_{2}^{\frac{1}{x}}

\int \frac{t}{t ^{2} + 1} d t = \frac{1}{2} ln t^{2} + 1

= [t arctan (t) - \frac{1}{2} ln t^{2} + 1]_{2}^{\frac{1}{x}}

Compute the boundaries:

\frac{1}{x} arctan (\frac{1}{x}) - \frac{1}{2} ln \frac{1}{x ^{2}} + 1 - 2 arctan (2) + \frac{1}{2} ln (5)

= \frac{1}{x} arctan (\frac{1}{x}) - \frac{1}{2} ln \frac{1}{x ^{2}} + 1 - 2 arctan (2) + \frac{1}{2} ln (5)

Simplify

= \frac{arctan ( \frac{1}{x} )}{x} - \frac{1}{2} ln \frac{1}{x ^{2}} + 1 - 2 arctan (2) + \frac{1}{2} ln (5)

(cos (π))^{^{'}}

\int_{- 5}^{0} (4 + 25 - x^{2}) d x

x \to 8 lim (\frac{x + 9}{x - 8})

y^{^{''}} + 256 y = 7 cos (5 t)

\frac{10}{s ^{2} + 25} + \frac{4}{s - 3}

What is the integral from 2 to 1/x of arctan(t) ?
The integral from 2 to 1/x of arctan(t) is (arctan(1/x))/x-1/2 ln| 1/(x^2)+1|-2arctan(2)+1/2 ln(5)