integral of (r(r^2-1))/(r^2+1)

解答

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

- ln r^{2} + 1 + \frac{1}{2} r^{2} + C

求解步骤

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

使用换元积分法

= \int \frac{u - 1}{2 ( u + 1 )} d u

提出常数:

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

= \frac{1}{2} \cdot \int \frac{u - 1}{u + 1} d u

\frac{u - 1}{u + 1} = - \frac{2}{u + 1} + 1

= \frac{1}{2} \cdot \int - \frac{2}{u + 1} + 1 d u

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (- \int \frac{2}{u + 1} d u + \int 1 d u)

\int \frac{2}{u + 1} d u = 2 ln ∣ u + 1 ∣

\int 1 d u = u

= \frac{1}{2} (- 2 ln ∣ u + 1 ∣ + u)

u = r^{2}

代回

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

化简

\frac{1}{2} (- 2 ln r^{2} + 1 + r^{2}) : - ln r^{2} + 1 + \frac{1}{2} r^{2}

= - ln r^{2} + 1 + \frac{1}{2} r^{2}

解答补常数

= - ln r^{2} + 1 + \frac{1}{2} r^{2} + C

\int \frac{csc ^{4} ( t )}{cot ^{2} ( t )} d t

\int cot^{5} (2 y) d y

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

\int \frac{( 1 + tan ( t ) ) ^{2}}{sec ^{2} ( t )} d t

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