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受欢迎的微积分 >

integral of (x^2)/((16+x^2)^{1/2)}

解答

\int \frac{x ^{2}}{( 16 + x ^{2} ) ^{\frac{1}{2}}} d x

解答

16 - \frac{1}{2} ln \frac{x + ( 16 + x ^{2} ) ^{\frac{1}{2}}}{4} + \frac{1}{32} x (16 + x^{2})^{\frac{1}{2}} + C

求解步骤

\int \frac{x ^{2}}{( 16 + x ^{2} ) ^{\frac{1}{2}}} d x

使用指数法则:

\frac{1}{a ^{b}} = a^{- b}

\frac{x ^{2}}{( 16 + x ^{2} ) ^{\frac{1}{2}}} = (x^{2}) (16 + x^{2})^{- \frac{1}{2}}

= \int x^{2} (16 + x^{2})^{- \frac{1}{2}} d x

使用三角换元法

= \int 16 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

= 16 \cdot \int tan^{2} (u) sec (u) d u

使用三角恒等式改写

= 16 \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

乘开

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

= 16 \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

= 16 (- \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) ∣

= 16 (- ln ∣ tan (u) + sec (u) ∣ + \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

u = arctan (\frac{1}{4} x)

代回

= 16 (- ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x)) + \frac{1}{2} sec (arctan (\frac{1}{4} x)) tan (arctan (\frac{1}{4} x)) + \frac{1}{2} ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x)))

化简

16 (- ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x)) + \frac{1}{2} sec (arctan (\frac{1}{4} x)) tan (arctan (\frac{1}{4} x)) + \frac{1}{2} ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x))) : 16 - \frac{1}{2} ln \frac{x + ( 16 + x ^{2} ) ^{\frac{1}{2}}}{4} + \frac{1}{32} x (16 + x^{2})^{\frac{1}{2}}

= 16 - \frac{1}{2} ln \frac{x + ( 16 + x ^{2} ) ^{\frac{1}{2}}}{4} + \frac{1}{32} x (16 + x^{2})^{\frac{1}{2}}

解答补常数

= 16 - \frac{1}{2} ln \frac{x + ( 16 + x ^{2} ) ^{\frac{1}{2}}}{4} + \frac{1}{32} x (16 + x^{2})^{\frac{1}{2}} + C

作图

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