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受欢迎的微积分 >

integral of (t^2)/(sqrt(a^4+t^2))

解答

\int \frac{t ^{2}}{a ^{4} + t ^{2}} d t

解答

\frac{1}{2} t t^{2} + a^{4} - \frac{a ^{4}}{2} (ln t + t^{2} + a^{4} - 2 ln (a)) + C

求解步骤

\int \frac{t ^{2}}{a ^{4} + t ^{2}} d t

使用三角换元法

= \int a^{4} tan^{2} (u) sec (u) d u

提出常数:

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

= a^{4} \cdot \int tan^{2} (u) sec (u) d u

使用三角恒等式改写

= a^{4} \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

乘开

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

= a^{4} \cdot \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

= a^{4} (- \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) ∣

= a^{4} (- ln ∣ tan (u) + sec (u) ∣ + \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

u = arctan (\frac{1}{a ^{2}} t)

代回

= a^{4} (- ln tan (arctan (\frac{1}{a ^{2}} t)) + sec (arctan (\frac{1}{a ^{2}} t)) + \frac{1}{2} sec (arctan (\frac{1}{a ^{2}} t)) tan (arctan (\frac{1}{a ^{2}} t)) + \frac{1}{2} ln tan (arctan (\frac{1}{a ^{2}} t)) + sec (arctan (\frac{1}{a ^{2}} t)))

化简

a^{4} (- ln tan (arctan (\frac{1}{a ^{2}} t)) + sec (arctan (\frac{1}{a ^{2}} t)) + \frac{1}{2} sec (arctan (\frac{1}{a ^{2}} t)) tan (arctan (\frac{1}{a ^{2}} t)) + \frac{1}{2} ln tan (arctan (\frac{1}{a ^{2}} t)) + sec (arctan (\frac{1}{a ^{2}} t))) : \frac{1}{2} t t^{2} + a^{4} - \frac{a ^{4}}{2} (ln t + t^{2} + a^{4} - 2 ln (a))

= \frac{1}{2} t t^{2} + a^{4} - \frac{a ^{4}}{2} (ln t + t^{2} + a^{4} - 2 ln (a))

解答补常数

= \frac{1}{2} t t^{2} + a^{4} - \frac{a ^{4}}{2} (ln t + t^{2} + a^{4} - 2 ln (a)) + C

作图

流行的例子

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\frac{d}{d x} (\frac{5 x}{6})

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\int 2^{x} + 4 sinh (x) d x

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