integral of tln(t+3)

解答

\int t ln (t + 3) d t

ln (t + 3) (\frac{1}{2} (t + 3)^{2} - 3 (t + 3)) - \frac{1}{4} (t^{2} - 6 t - 27) + C

求解步骤

\int t ln (t + 3) d t

使用换元积分法

= \int (u - 3) ln (u) d u

使用分布积分法

= ln (u) (\frac{u ^{2}}{2} - 3 u) - \int \frac{1}{2} (u - 6) d u

\int \frac{1}{2} (u - 6) d u = \frac{1}{2} (\frac{u ^{2}}{2} - 6 u)

= ln (u) (\frac{u ^{2}}{2} - 3 u) - \frac{1}{2} (\frac{u ^{2}}{2} - 6 u)

u = t + 3

代回

= ln (t + 3) (\frac{( t + 3 ) ^{2}}{2} - 3 (t + 3)) - \frac{1}{2} (\frac{( t + 3 ) ^{2}}{2} - 6 (t + 3))

化简

ln (t + 3) (\frac{( t + 3 ) ^{2}}{2} - 3 (t + 3)) - \frac{1}{2} (\frac{( t + 3 ) ^{2}}{2} - 6 (t + 3)) : ln (t + 3) (\frac{1}{2} (t + 3)^{2} - 3 (t + 3)) - \frac{1}{4} (t^{2} - 6 t - 27)

= ln (t + 3) (\frac{1}{2} (t + 3)^{2} - 3 (t + 3)) - \frac{1}{4} (t^{2} - 6 t - 27)

解答补常数

= ln (t + 3) (\frac{1}{2} (t + 3)^{2} - 3 (t + 3)) - \frac{1}{4} (t^{2} - 6 t - 27) + C

\int \frac{1}{b} sin (b x) d x

\int \frac{2 ln ( x )}{x ^{2}} d x

n or ma l y = x^{2} - 9 x, (7, - 14)

\int_{- 1}^{1} 3 x d x

x \to 0 + lim (x^{15} ln (x))