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受欢迎的微积分 >

integral of 1/(sqrt(2x^2+x))

解答

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

解答

\frac{1}{2} ln 16 x^{2} + 8 x + 4 x + 1 + C

求解步骤

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

对

2 x^{2} + x

配方:

2 (x + \frac{1}{4})^{2} - \frac{1}{8}

= \int \frac{1}{2 ( x + \frac{1}{4} ) ^{2} - \frac{1}{8}} d x

使用换元积分法

= \int \frac{2 2}{16 u ^{2} - 1} d u

提出常数:

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

= 22 \cdot \int \frac{1}{16 u ^{2} - 1} d u

使用三角换元法

= 22 \cdot \int \frac{sec ( v )}{4} d v

提出常数:

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

= 22 \frac{1}{4} \cdot \int sec (v) d v

使用积分常见定则:

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

= 22 \frac{1}{4} ln ∣ tan (v) + sec (v) ∣

代回

= 22 \frac{1}{4} ln tan (arcsec (4 (x + \frac{1}{4}))) + sec (arcsec (4 (x + \frac{1}{4})))

化简

22 \frac{1}{4} ln tan (arcsec (4 (x + \frac{1}{4}))) + sec (arcsec (4 (x + \frac{1}{4}))) : \frac{1}{2} ln 16 x^{2} + 8 x + 4 x + 1

= \frac{1}{2} ln 16 x^{2} + 8 x + 4 x + 1

解答补常数

= \frac{1}{2} ln 16 x^{2} + 8 x + 4 x + 1 + C

作图

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\frac{d}{d x} ((cos (x))^{s i n (x)})

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a re a e^{x}, 14 e^{- x} + 5, x = 0

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\int x (4 x^{2} + 8) d x

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\int x 1 - x^{4} d x

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