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受欢迎的微积分 >

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

解答

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

解答

\frac{1}{4} (ln ∣ 1 + 2 x ∣ + \frac{1}{1 + 2 x}) + C

求解步骤

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

使用换元积分法

= \int \frac{u - 1}{4 u ^{2}} d u

提出常数:

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

= \frac{1}{4} \cdot \int \frac{u - 1}{u ^{2}} d u

乘开

\frac{u - 1}{u ^{2}} : \frac{1}{u} - \frac{1}{u ^{2}}

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

= \frac{1}{4} (ln ∣ u ∣ - (- \frac{1}{u}))

u = 1 + 2 x

代回

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

化简

= \frac{1}{4} (ln ∣ 1 + 2 x ∣ + \frac{1}{1 + 2 x})

解答补常数

= \frac{1}{4} (ln ∣ 1 + 2 x ∣ + \frac{1}{1 + 2 x}) + C

作图

流行的例子

(\partial)/(\partial t)(2s^2t^2)

\frac{\partial}{\partial t} (2 s^{2} t^{2})

integral of x^2sqrt(8+x)

\int x^{2} 8 + x d x

integral of 7/(9+4x^2)

\int \frac{7}{9 + 4 x ^{2}} d x

derivative of ln(x/7)

\frac{d}{d x} (ln (\frac{x}{7}))

y^{''}+5y^'+6=1

y^{^{''}} + 5 y^{^{'}} + 6 = 1