What is the integral of 1/((x+1)^2)arctan(x) ?

The integral of 1/((x+1)^2)arctan(x) is -(arctan(x))/(x+1)-1/4 ln|x^2+1|+1/2 arctan(x)+1/2 ln|x+1|+C

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integral of 1/((x+1)^2)arctan(x)

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

- \frac{arctan ( x )}{x + 1} - \frac{1}{4} ln x^{2} + 1 + \frac{1}{2} arctan (x) + \frac{1}{2} ln ∣ x + 1 ∣ + C

Solution steps

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

Simplify

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

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

Apply Integration By Parts

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

\int - \frac{1}{( x ^{2} + 1 ) ( x + 1 )} d x = \frac{1}{4} ln x^{2} + 1 - \frac{1}{2} arctan (x) - \frac{1}{2} ln ∣ x + 1 ∣

= - \frac{arctan ( x )}{x + 1} - (\frac{1}{4} ln x^{2} + 1 - \frac{1}{2} arctan (x) - \frac{1}{2} ln ∣ x + 1 ∣)

Simplify

= - \frac{arctan ( x )}{x + 1} - \frac{1}{4} ln x^{2} + 1 + \frac{1}{2} arctan (x) + \frac{1}{2} ln ∣ x + 1 ∣

Add a constant to the solution

= - \frac{arctan ( x )}{x + 1} - \frac{1}{4} ln x^{2} + 1 + \frac{1}{2} arctan (x) + \frac{1}{2} ln ∣ x + 1 ∣ + C

What is the integral of 1/((x+1)^2)arctan(x) ?
The integral of 1/((x+1)^2)arctan(x) is -(arctan(x))/(x+1)-1/4 ln|x^2+1|+1/2 arctan(x)+1/2 ln|x+1|+C