integral of+tln(t+5)

解答

\int + t ln (t + 5) d t

ln (t + 5) (\frac{1}{2} (t + 5)^{2} - 5 (t + 5)) - \frac{1}{4} (t^{2} - 10 t - 75) + C

求解步骤

\int t ln (t + 5) d t

使用换元积分法

= \int (u - 5) ln (u) d u

使用分布积分法

= ln (u) (\frac{u ^{2}}{2} - 5 u) - \int \frac{1}{2} (u - 10) d u

\int \frac{1}{2} (u - 10) d u = \frac{1}{2} (\frac{u ^{2}}{2} - 10 u)

= ln (u) (\frac{u ^{2}}{2} - 5 u) - \frac{1}{2} (\frac{u ^{2}}{2} - 10 u)

u = t + 5

代回

= ln (t + 5) (\frac{( t + 5 ) ^{2}}{2} - 5 (t + 5)) - \frac{1}{2} (\frac{( t + 5 ) ^{2}}{2} - 10 (t + 5))

化简

ln (t + 5) (\frac{( t + 5 ) ^{2}}{2} - 5 (t + 5)) - \frac{1}{2} (\frac{( t + 5 ) ^{2}}{2} - 10 (t + 5)) : ln (t + 5) (\frac{1}{2} (t + 5)^{2} - 5 (t + 5)) - \frac{1}{4} (t^{2} - 10 t - 75)

= ln (t + 5) (\frac{1}{2} (t + 5)^{2} - 5 (t + 5)) - \frac{1}{4} (t^{2} - 10 t - 75)

解答补常数

= ln (t + 5) (\frac{1}{2} (t + 5)^{2} - 5 (t + 5)) - \frac{1}{4} (t^{2} - 10 t - 75) + C

\frac{d}{d x} (\frac{3}{2 + tan ( x )})

\int \frac{tan ( x )}{sin ( x ) cos ( x )} d x

\frac{d y}{d x} = \frac{2 sec ( y )}{( x + 1 ) ^{2}}

\int \frac{( cos ( x ) )}{3 - sin ( x )} d x

f (x) = \frac{e ^{x} + e ^{- x}}{2}