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受欢迎的微积分 >

integral of-(t^2)/(sqrt(t^2+4t+3))

解答

\int - \frac{t ^{2}}{t ^{2} + 4 t + 3} d t

解答

- \frac{1}{2} (t + 2) t^{2} + 4 t + 3 - \frac{1}{2} ln t^{2} + 4 t + 3 + t + 2 + 4 t^{2} + 4 t + 3 - 4 ln t^{2} + 4 t + 3 + t + 2 + C

求解步骤

\int - \frac{t ^{2}}{t ^{2} + 4 t + 3} d t

提出常数:

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

= - \int \frac{t ^{2}}{t ^{2} + 4 t + 3} d t

对

t^{2} + 4 t + 3

配方:

(t + 2)^{2} - 1

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

使用换元积分法

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

乘开

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

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

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

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

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

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

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

u = t + 2

代回

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

化简

- (\frac{1}{2} (t + 2) (t + 2)^{2} - 1 + \frac{1}{2} ln (t + 2)^{2} - 1 + t + 2 - 4 (t + 2)^{2} - 1 + 4 ln (t + 2)^{2} - 1 + t + 2) : - \frac{1}{2} (t + 2) t^{2} + 4 t + 3 - \frac{1}{2} ln t^{2} + 4 t + 3 + t + 2 + 4 t^{2} + 4 t + 3 - 4 ln t^{2} + 4 t + 3 + t + 2

= - \frac{1}{2} (t + 2) t^{2} + 4 t + 3 - \frac{1}{2} ln t^{2} + 4 t + 3 + t + 2 + 4 t^{2} + 4 t + 3 - 4 ln t^{2} + 4 t + 3 + t + 2

解答补常数

= - \frac{1}{2} (t + 2) t^{2} + 4 t + 3 - \frac{1}{2} ln t^{2} + 4 t + 3 + t + 2 + 4 t^{2} + 4 t + 3 - 4 ln t^{2} + 4 t + 3 + t + 2 + C

作图

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