integral of (10)/(4x^3-4x^2+5x)

解答

\int \frac{10}{4 x ^{3} - 4 x ^{2} + 5 x} d x

2 ln ∣ x ∣ - ln 4 x^{2} - 4 x + 5 + arctan (x - \frac{1}{2}) + C

求解步骤

\int \frac{10}{4 x ^{3} - 4 x ^{2} + 5 x} d x

提出常数:

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

= 10 \cdot \int \frac{1}{4 x ^{3} - 4 x ^{2} + 5 x} d x

将

\frac{1}{4 x ^{3} - 4 x ^{2} + 5 x}

用部份分式展开:

\frac{1}{5 x} + \frac{- 4 x + 4}{5 ( 4 x ^{2} - 4 x + 5 )}

= 10 \cdot \int \frac{1}{5 x} + \frac{- 4 x + 4}{5 ( 4 x ^{2} - 4 x + 5 )} d x

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

= 10 (\int \frac{1}{5 x} d x + \int \frac{- 4 x + 4}{5 ( 4 x ^{2} - 4 x + 5 )} d x)

\int \frac{1}{5 x} d x = \frac{1}{5} ln ∣ x ∣

\int \frac{- 4 x + 4}{5 ( 4 x ^{2} - 4 x + 5 )} d x = \frac{1}{10} (- ln 4 x^{2} - 4 x + 5 + arctan (x - \frac{1}{2}))

= 10 (\frac{1}{5} ln ∣ x ∣ + \frac{1}{10} (- ln 4 x^{2} - 4 x + 5 + arctan (x - \frac{1}{2})))

化简

10 (\frac{1}{5} ln ∣ x ∣ + \frac{1}{10} (- ln 4 x^{2} - 4 x + 5 + arctan (x - \frac{1}{2}))) : 2 ln ∣ x ∣ - ln 4 x^{2} - 4 x + 5 + arctan (x - \frac{1}{2})

= 2 ln ∣ x ∣ - ln 4 x^{2} - 4 x + 5 + arctan (x - \frac{1}{2})

解答补常数

= 2 ln ∣ x ∣ - ln 4 x^{2} - 4 x + 5 + arctan (x - \frac{1}{2}) + C

\frac{d}{d x} (\frac{1}{4} x^{2} - 2 x - y)

\frac{\partial}{\partial y} (ln (x y))

\int x^{2} 3 - 2 x d x

\int \frac{1}{( a ^{2} + v ^{2} )} d v

y^{^{'}} + tan (x) y - sin (x) = 0