What is the integral of arctan(1/t) ?

The integral of arctan(1/t) is tarctan(1/t)+1/2 ln|1+t^2|+C

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integral of arctan(1/t)

\int arctan (\frac{1}{t}) d t

t arctan (\frac{1}{t}) + \frac{1}{2} ln 1 + t^{2} + C

Solution steps

\int arctan (\frac{1}{t}) d t

Apply Integration By Parts

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

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

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

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

Simplify

= t arctan (\frac{1}{t}) + \frac{1}{2} ln 1 + t^{2}

Add a constant to the solution

= t arctan (\frac{1}{t}) + \frac{1}{2} ln 1 + t^{2} + C

\int \frac{3 x + 2}{x ^{2} - 4 x + 20} d x

e^{\frac{x}{2}}

\int \frac{2 x ^{2} - x + 2}{2 x - 3} d x

\int \frac{4 ^{x}}{ln ( 4 )} d x

\int 4 x - 10 x e^{- 0.2 x} d x

What is the integral of arctan(1/t) ?
The integral of arctan(1/t) is tarctan(1/t)+1/2 ln|1+t^2|+C