integral of tln(t+1)

解答

\int t ln (t + 1) d t

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

求解步骤

\int t ln (t + 1) d t

使用换元积分法

= \int (u - 1) ln (u) d u

使用分布积分法

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

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

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

u = t + 1

代回

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

化简

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

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

解答补常数

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

\frac{\partial}{\partial x} (2 (1 + y) x)

d er i v a t i v e y = 3^{x^{2} + 3 x}

\frac{d}{d y} (\frac{y ^{3}}{27} + \frac{9}{4 y})

d er i v a t i v e sin (4 x) + sin^{2} (4 x)

f (x) = - 3 lo g_{10} (x)