integral of (x^2ln(x))/((x^3+1)^2)

解答

\int \frac{x ^{2} ln ( x )}{( x ^{3} + 1 ) ^{2}} d x

\frac{1}{9} (- \frac{3 ln ( x )}{x ^{3} + 1} + ln x^{3} - ln x^{3} + 1) + C

求解步骤

\int \frac{x ^{2} ln ( x )}{( x ^{3} + 1 ) ^{2}} d x

使用换元积分法

= \int \frac{ln ( u )}{9 ( u + 1 ) ^{2}} d u

提出常数:

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

= \frac{1}{9} \cdot \int \frac{ln ( u )}{( u + 1 ) ^{2}} d u

使用分布积分法

= \frac{1}{9} (- \frac{ln ( u )}{u + 1} - \int - \frac{1}{u ( u + 1 )} d u)

\int - \frac{1}{u ( u + 1 )} d u = - ln ∣ u ∣ + ln ∣ u + 1 ∣

= \frac{1}{9} (- \frac{ln ( u )}{u + 1} - (- ln ∣ u ∣ + ln ∣ u + 1 ∣))

u = x^{3}

代回

= \frac{1}{9} (- \frac{ln ( x ^{3} )}{x ^{3} + 1} - (- ln x^{3} + ln x^{3} + 1))

化简

\frac{1}{9} (- \frac{ln ( x ^{3} )}{x ^{3} + 1} - (- ln x^{3} + ln x^{3} + 1)) : \frac{1}{9} (- \frac{3 ln ( x )}{x ^{3} + 1} + ln x^{3} - ln x^{3} + 1)

= \frac{1}{9} (- \frac{3 ln ( x )}{x ^{3} + 1} + ln x^{3} - ln x^{3} + 1)

解答补常数

= \frac{1}{9} (- \frac{3 ln ( x )}{x ^{3} + 1} + ln x^{3} - ln x^{3} + 1) + C

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

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

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

\int \frac{3 ln ^{3} ( x ) + 5 ln ( x ^{2} )}{x} d x

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