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integral of (t^3+5t^2+7t+1)/(t^2+3t)

解答

\int \frac{t ^{3} + 5 t ^{2} + 7 t + 1}{t ^{2} + 3 t} d t

解答

\frac{t ^{2}}{2} + 2 t + \frac{1}{3} ln ∣ t ∣ + \frac{2}{3} ln ∣ t + 3 ∣ + C

求解步骤

\int \frac{t ^{3} + 5 t ^{2} + 7 t + 1}{t ^{2} + 3 t} d t

对

t^{2} + 3 t

配方:

(t + \frac{3}{2})^{2} - \frac{9}{4}

= \int \frac{t ^{3} + 5 t ^{2} + 7 t + 1}{( t + \frac{3}{2} ) ^{2} - \frac{9}{4}} d t

使用换元积分法

= \int \frac{20 u ^{2} - 32 u + 4 ( u - \frac{3}{2} ) ^{3} + 7}{4 u ^{2} - 9} d u

使用换元积分法

= \int \frac{v ^{3} + 5 v ^{2} + 7 v + 1}{v ( v + 3 )} d v

将

\frac{v ^{3} + 5 v ^{2} + 7 v + 1}{v ( v + 3 )}

用部份分式展开:

v + 2 + \frac{1}{3 v} + \frac{2}{3 ( v + 3 )}

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

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

\int v d v = \frac{v ^{2}}{2}

\int 2 d v = 2 v

\int \frac{1}{3 v} d v = \frac{1}{3} ln ∣ v ∣

\int \frac{2}{3 ( v + 3 )} d v = \frac{2}{3} ln ∣ v + 3 ∣

= \frac{v ^{2}}{2} + 2 v + \frac{1}{3} ln ∣ v ∣ + \frac{2}{3} ln ∣ v + 3 ∣

代回

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

化简

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

= \frac{t ^{2}}{2} + 2 t + \frac{1}{3} ln ∣ t ∣ + \frac{2}{3} ln ∣ t + 3 ∣

解答补常数

= \frac{t ^{2}}{2} + 2 t + \frac{1}{3} ln ∣ t ∣ + \frac{2}{3} ln ∣ t + 3 ∣ + C

作图

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