zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的微积分 >

integral of 8/(t(t^2+2)^2)

解答

\int \frac{8}{t ( t ^{2} + 2 ) ^{2}} d t

解答

2 ln (t) - ln t^{2} + 2 + \frac{2}{t ^{2} + 2} + C

求解步骤

\int \frac{8}{t ( t ^{2} + 2 ) ^{2}} d t

提出常数:

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

= 8 \cdot \int \frac{1}{t ( t ^{2} + 2 ) ^{2}} d t

使用换元积分法

= 8 \cdot \int \frac{1}{2 u ( u + 2 ) ^{2}} d u

提出常数:

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

= 8 \cdot \frac{1}{2} \cdot \int \frac{1}{u ( u + 2 ) ^{2}} d u

将

\frac{1}{u ( u + 2 ) ^{2}}

用部份分式展开:

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

= 8 \cdot \frac{1}{2} \cdot \int \frac{1}{4 u} - \frac{1}{4 ( u + 2 )} - \frac{1}{2 ( u + 2 ) ^{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

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

\int \frac{1}{4 u} d u = \frac{1}{4} ln ∣ u ∣

\int \frac{1}{4 ( u + 2 )} d u = \frac{1}{4} ln ∣ u + 2 ∣

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

= 8 \cdot \frac{1}{2} (\frac{1}{4} ln ∣ u ∣ - \frac{1}{4} ln ∣ u + 2 ∣ - (- \frac{1}{2 ( u + 2 )}))

u = t^{2}

代回

= 8 \cdot \frac{1}{2} (\frac{1}{4} ln t^{2} - \frac{1}{4} ln t^{2} + 2 - (- \frac{1}{2 ( t ^{2} + 2 )}))

化简

8 \cdot \frac{1}{2} (\frac{1}{4} ln t^{2} - \frac{1}{4} ln t^{2} + 2 - (- \frac{1}{2 ( t ^{2} + 2 )})) : 2 ln (t) - ln t^{2} + 2 + \frac{2}{t ^{2} + 2}

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

解答补常数

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

作图

流行的例子

integral of \sqrt[3]{(x^2+1)^4}x

\int 3 (x^{2} + 1)^{4} x d x

integral of sqrt(16x^2+25)

\int 16 x^{2} + 25 d x

integral of ((x^7+2))/((x^2+x+1)^2)

\int \frac{( x ^{7} + 2 )}{( x ^{2} + x + 1 ) ^{2}} d x

integral of (5x^2+2x+29)/((x-3)(x^2+7))

\int \frac{5 x ^{2} + 2 x + 29}{( x - 3 ) ( x ^{2} + 7 )} d x

integral of 7tan^4(x)sec^4(x)

\int 7 tan^{4} (x) sec^{4} (x) d x