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受欢迎的微积分 >

integral of 1/((x+1)^2+3)

解答

\int \frac{1}{( x + 1 ) ^{2} + 3} d x

解答

\frac{1}{3} arctan (\frac{1}{3} (x + 1)) + C

求解步骤

\int \frac{1}{( x + 1 ) ^{2} + 3} d x

(x + 1)^{2} + 3 = 3 (\frac{( x + 1 ) ^{2}}{3} + 1)

= \int \frac{1}{3 ( \frac{( x + 1 ) ^{2}}{3} + 1 )} d x

使用指数法则:

\frac{a ^{n}}{b ^{n}} = (\frac{a}{b})^{n}

\frac{( x + 1 ) ^{2}}{3} = (\frac{1}{3} (x + 1))^{2}

= \int \frac{1}{3 ( ( \frac{1}{3} ( x + 1 ) ) ^{2} + 1 )} d x

提出常数:

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

= \frac{1}{3} \cdot \int \frac{1}{( \frac{1}{3} ( x + 1 ) ) ^{2} + 1} d x

化简

\frac{1}{( \frac{1}{3} ( x + 1 ) ) ^{2} + 1} : \frac{3}{x ^{2} + 2 x + 4}

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

提出常数:

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

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

对

x^{2} + 2 x + 4

配方:

(x + 1)^{2} + 3

= \frac{1}{3} \cdot 3 \cdot \int \frac{1}{( x + 1 ) ^{2} + 3} d x

使用换元积分法

= \frac{1}{3} \cdot 3 \cdot \int \frac{1}{u ^{2} + 3} d u

使用换元积分法

= \frac{1}{3} \cdot 3 \cdot \int \frac{1}{3 ( v ^{2} + 1 )} d v

提出常数:

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

= \frac{1}{3} \cdot 3 \cdot \frac{1}{3} \cdot \int \frac{1}{v ^{2} + 1} d v

使用积分常见定则:

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

= \frac{1}{3} \cdot 3 \cdot \frac{1}{3} arctan (v)

代回

= \frac{1}{3} \cdot 3 \cdot \frac{1}{3} arctan (\frac{x + 1}{3})

化简

\frac{1}{3} \cdot 3 \cdot \frac{1}{3} arctan (\frac{x + 1}{3}) : \frac{1}{3} arctan (\frac{1}{3} (x + 1))

= \frac{1}{3} arctan (\frac{1}{3} (x + 1))

解答补常数

= \frac{1}{3} arctan (\frac{1}{3} (x + 1)) + C

作图

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