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受欢迎的微积分 >

integral of (t^5)/(sqrt(t^2+2))

解答

\int \frac{t ^{5}}{t ^{2} + 2} d t

解答

42 \frac{1}{2} (2 + t^{2}) - \frac{4}{3} (2 + t^{2})^{\frac{3}{2}} + \frac{1}{5} (2 + t^{2})^{\frac{5}{2}} + C

求解步骤

\int \frac{t ^{5}}{t ^{2} + 2} d t

使用三角换元法

= \int 42 tan^{5} (u) sec (u) d u

提出常数:

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

= 42 \cdot \int tan^{5} (u) sec (u) d u

使用三角恒等式改写

= 42 \cdot \int (- 1 + sec^{2} (u))^{2} tan (u) sec (u) d u

使用换元积分法

= 42 \cdot \int (- 1 + v^{2})^{2} d v

乘开

(- 1 + v^{2})^{2} : 1 - 2 v^{2} + v^{4}

= 42 \cdot \int 1 - 2 v^{2} + v^{4} 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

= 42 (\int 1 d v - \int 2 v^{2} d v + \int v^{4} d v)

\int 1 d v = v

\int 2 v^{2} d v = \frac{2 v ^{3}}{3}

\int v^{4} d v = \frac{v ^{5}}{5}

= 42 (v - \frac{2 v ^{3}}{3} + \frac{v ^{5}}{5})

代回

= 42 sec (arctan (\frac{1}{2} t)) - \frac{2 sec ^{3} ( arctan ( \frac{1}{2} t ) )}{3} + \frac{sec ^{5} ( arctan ( \frac{1}{2} t ) )}{5}

化简

42 sec (arctan (\frac{1}{2} t)) - \frac{2 sec ^{3} ( arctan ( \frac{1}{2} t ) )}{3} + \frac{sec ^{5} ( arctan ( \frac{1}{2} t ) )}{5} : 42 \frac{1}{2} (2 + t^{2}) - \frac{4}{3} (2 + t^{2})^{\frac{3}{2}} + \frac{1}{5} (2 + t^{2})^{\frac{5}{2}}

= 42 \frac{1}{2} (2 + t^{2}) - \frac{4}{3} (2 + t^{2})^{\frac{3}{2}} + \frac{1}{5} (2 + t^{2})^{\frac{5}{2}}

解答补常数

= 42 \frac{1}{2} (2 + t^{2}) - \frac{4}{3} (2 + t^{2})^{\frac{3}{2}} + \frac{1}{5} (2 + t^{2})^{\frac{5}{2}} + C

作图

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