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

解答

\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

求解步骤

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

化简

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

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

使用分布积分法

= - \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 ∣)

化简

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

解答补常数

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

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