What is the integral of 1rctan(x) ?

The integral of 1rctan(x) is d(xarctan(x)-1/2 ln|x^2+1|)+C

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integral of 1rctan(x)

\int 1 d arctan (x) d x

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

Solution steps

\int 1 \cdot d arctan (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 1 \cdot d \cdot \int arctan (x) d x

Simplify

= d \cdot \int arctan (x) d x

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

\int x^{2} (a^{2} - x^{2}) d x

\int 2 x^{2} - x + 1 d x

\int \frac{1}{P} d P

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

\int \frac{1 - x}{x + 1} d x

What is the integral of 1rctan(x) ?
The integral of 1rctan(x) is d(xarctan(x)-1/2 ln|x^2+1|)+C