integral of t(t^2+1)^5

解答

\int t (t^{2} + 1)^{5} d t

\frac{1}{12} (t^{2} + 1)^{6} + C

求解步骤

\int t (t^{2} + 1)^{5} d t

使用换元积分法

= \int \frac{u ^{5}}{2} d u

提出常数:

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

= \frac{1}{2} \cdot \int u^{5} d u

使用幂法则

= \frac{1}{2} \cdot \frac{u ^{6}}{6}

u = t^{2} + 1

代回

= \frac{1}{2} \cdot \frac{( t ^{2} + 1 ) ^{6}}{6}

化简

\frac{1}{2} \cdot \frac{( t ^{2} + 1 ) ^{6}}{6} : \frac{1}{12} (t^{2} + 1)^{6}

= \frac{1}{12} (t^{2} + 1)^{6}

解答补常数

= \frac{1}{12} (t^{2} + 1)^{6} + C

n = 0 \sum \infty \frac{n - 1}{n ^{2}}

\int 1 + e^{t} d t

y^{^{''}} + 0.1 y^{^{'}} + 0.0004 y = 0

t an g e n t f (x) = \frac{x - 6}{x + 2}, at x = 2

d er i v a t i v e 8 x - 3 x