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受欢迎的微积分 >

integral of (2t)/(1-t^2)

解答

\int \frac{2 t}{1 - t ^{2}} d t

解答

- ln 1 - t^{2} + C

求解步骤

\int \frac{2 t}{1 - t ^{2}} d t

提出常数:

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

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

使用换元积分法

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

提出常数:

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

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

使用积分常见定则:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 (- \frac{1}{2} ln ∣ u ∣)

u = 1 - t^{2}

代回

= 2 (- \frac{1}{2} ln 1 - t^{2})

化简

2 (- \frac{1}{2} ln 1 - t^{2}) : - ln 1 - t^{2}

= - ln 1 - t^{2}

解答补常数

= - ln 1 - t^{2} + C

作图

流行的例子

sum from n=1 to infinity of n*2^{-n}

n = 1 \sum \infty n \cdot 2^{- n}

-5t^2y^{''}-2ty^'+14y=0

- 5 t^{2} y^{^{''}} - 2 t y^{^{'}} + 14 y = 0

derivative x^2e^x-2xe^x+5e^x

d er i v a t i v e x^{2} e^{x} - 2 x e^{x} + 5 e^{x}

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x \to 0 lim (\frac{x ^{2} + 5 x}{x})

x y^{^{'}} = y + x