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integral of (u^2)/(sqrt(u^2+25))

解答

\int \frac{u ^{2}}{u ^{2} + 25} d u

解答

25 (- \frac{1}{2} ln (\frac{u + 25 + u ^{2}}{5}) + \frac{1}{50} u 25 + u^{2}) + C

求解步骤

\int \frac{u ^{2}}{u ^{2} + 25} d u

使用三角换元法

= \int 25 tan^{2} (v) sec (v) d v

提出常数:

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

= 25 \cdot \int tan^{2} (v) sec (v) d v

使用三角恒等式改写

= 25 \cdot \int (- 1 + sec^{2} (v)) sec (v) d v

乘开

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

= 25 \cdot \int - sec (v) + sec^{3} (v) d v

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

= 25 (- \int sec (v) d v + \int sec^{3} (v) d v)

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

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

= 25 (- ln ∣ tan (v) + sec (v) ∣ + \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

v = arctan (\frac{1}{5} u)

代回

= 25 (- ln tan (arctan (\frac{1}{5} u)) + sec (arctan (\frac{1}{5} u)) + \frac{1}{2} sec (arctan (\frac{1}{5} u)) tan (arctan (\frac{1}{5} u)) + \frac{1}{2} ln tan (arctan (\frac{1}{5} u)) + sec (arctan (\frac{1}{5} u)))

化简

25 (- ln tan (arctan (\frac{1}{5} u)) + sec (arctan (\frac{1}{5} u)) + \frac{1}{2} sec (arctan (\frac{1}{5} u)) tan (arctan (\frac{1}{5} u)) + \frac{1}{2} ln tan (arctan (\frac{1}{5} u)) + sec (arctan (\frac{1}{5} u))) : 25 (- \frac{1}{2} ln (\frac{u + 25 + u ^{2}}{5}) + \frac{1}{50} u 25 + u^{2})

= 25 (- \frac{1}{2} ln (\frac{u + 25 + u ^{2}}{5}) + \frac{1}{50} u 25 + u^{2})

解答补常数

= 25 (- \frac{1}{2} ln (\frac{u + 25 + u ^{2}}{5}) + \frac{1}{50} u 25 + u^{2}) + C

作图

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