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受欢迎的微积分 >

积分 (r^2)/(sqrt(1+r^2))

解答

积分

\frac{r ^{2}}{1 + r ^{2}}

解答

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

求解步骤

\int \frac{r ^{2}}{1 + r ^{2}} d r

使用三角换元法

= \int tan^{2} (u) sec (u) d u

使用三角恒等式改写

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

乘开

(- 1 + sec^{2} (u)) sec (u) : - sec (u) + sec^{3} (u)

= \int - sec (u) + sec^{3} (u) 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

= - \int sec (u) d u + \int sec^{3} (u) d u

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

\int sec^{3} (u) d u = \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

= - ln ∣ tan (u) + sec (u) ∣ + \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

u = arctan (r)

代回

= - ln ∣ tan (arctan (r)) + sec (arctan (r)) ∣ + \frac{1}{2} sec (arctan (r)) tan (arctan (r)) + \frac{1}{2} ln ∣ tan (arctan (r)) + sec (arctan (r)) ∣

化简

- ln ∣ tan (arctan (r)) + sec (arctan (r)) ∣ + \frac{1}{2} sec (arctan (r)) tan (arctan (r)) + \frac{1}{2} ln ∣ tan (arctan (r)) + sec (arctan (r)) ∣ : \frac{1}{2} r r^{2} + 1 - \frac{1}{2} ln r + r^{2} + 1

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

解答补常数

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

作图

流行的例子

积分 1/((x^2+x+1)^2)

in t e g r a l \frac{1}{( x ^{2} + x + 1 ) ^{2}}

in t e g r a l \frac{x}{x ^{2}}

积分 1/(x^2+4x+9)

in t e g r a l \frac{1}{x ^{2} + 4 x + 9}

in t e g r a l \frac{1}{x ^{6}}

in t e g r a l 2 e^{u}