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受欢迎的微积分 >

integral of (x^2)/(sqrt(9x^2+4))

解答

\int \frac{x ^{2}}{9 x ^{2} + 4} d x

解答

\frac{1}{54} (- 4 ln (\frac{3 x + 9 x ^{2} + 4}{2}) + 3 x 9 x^{2} + 4) + C

求解步骤

\int \frac{x ^{2}}{9 x ^{2} + 4} d x

使用三角换元法

= \int \frac{4 tan ^{2} ( u ) sec ( u )}{27} d u

提出常数:

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

= \frac{4}{27} \cdot \int tan^{2} (u) sec (u) d u

使用三角恒等式改写

= \frac{4}{27} \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

乘开

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

= \frac{4}{27} \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

= \frac{4}{27} (- \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) ∣

= \frac{4}{27} (- ln ∣ tan (u) + sec (u) ∣ + \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

u = arctan (\frac{3}{2} x)

代回

= \frac{4}{27} (- ln tan (arctan (\frac{3}{2} x)) + sec (arctan (\frac{3}{2} x)) + \frac{1}{2} sec (arctan (\frac{3}{2} x)) tan (arctan (\frac{3}{2} x)) + \frac{1}{2} ln tan (arctan (\frac{3}{2} x)) + sec (arctan (\frac{3}{2} x)))

化简

\frac{4}{27} (- ln tan (arctan (\frac{3}{2} x)) + sec (arctan (\frac{3}{2} x)) + \frac{1}{2} sec (arctan (\frac{3}{2} x)) tan (arctan (\frac{3}{2} x)) + \frac{1}{2} ln tan (arctan (\frac{3}{2} x)) + sec (arctan (\frac{3}{2} x))) : \frac{1}{54} (- 4 ln (\frac{3 x + 9 x ^{2} + 4}{2}) + 3 x 9 x^{2} + 4)

= \frac{1}{54} (- 4 ln (\frac{3 x + 9 x ^{2} + 4}{2}) + 3 x 9 x^{2} + 4)

解答补常数

= \frac{1}{54} (- 4 ln (\frac{3 x + 9 x ^{2} + 4}{2}) + 3 x 9 x^{2} + 4) + C

作图

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