integral of t/(sqrt(1+3t))

解答

\int \frac{t}{1 + 3 t} d t

\frac{2}{9} (\frac{1}{3} (1 + 3 t)^{\frac{3}{2}} - 1 + 3 t) + C

求解步骤

\int \frac{t}{1 + 3 t} d t

使用换元积分法

= \int \frac{2 ( u ^{2} - 1 )}{9} d u

提出常数:

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

= \frac{2}{9} \cdot \int 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

= \frac{2}{9} (\int u^{2} d u - \int 1 d u)

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

\int 1 d u = u

= \frac{2}{9} (\frac{u ^{3}}{3} - u)

u = 1 + 3 t

代回

= \frac{2}{9} (\frac{( 1 + 3 t ) ^{3}}{3} - 1 + 3 t)

化简

\frac{2}{9} (\frac{( 1 + 3 t ) ^{3}}{3} - 1 + 3 t) : \frac{2}{9} (\frac{1}{3} (1 + 3 t)^{\frac{3}{2}} - 1 + 3 t)

= \frac{2}{9} (\frac{1}{3} (1 + 3 t)^{\frac{3}{2}} - 1 + 3 t)

解答补常数

= \frac{2}{9} (\frac{1}{3} (1 + 3 t)^{\frac{3}{2}} - 1 + 3 t) + C