integral of ((t+9)^2)/(t^2)

解答

\int \frac{( t + 9 ) ^{2}}{t ^{2}} d t

t + 18 ln ∣ t ∣ - \frac{81}{t} + C

求解步骤

\int \frac{( t + 9 ) ^{2}}{t ^{2}} d t

乘开

\frac{( t + 9 ) ^{2}}{t ^{2}} : 1 + \frac{18}{t} + \frac{81}{t ^{2}}

= \int 1 + \frac{18}{t} + \frac{81}{t ^{2}} d t

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

= \int 1 d t + \int \frac{18}{t} d t + \int \frac{81}{t ^{2}} d t

\int 1 d t = t

\int \frac{18}{t} d t = 18 ln ∣ t ∣

\int \frac{81}{t ^{2}} d t = - \frac{81}{t}

= t + 18 ln ∣ t ∣ - \frac{81}{t}

解答补常数

= t + 18 ln ∣ t ∣ - \frac{81}{t} + C

\int sec^{2} (x) cos^{2} (x) d x

\int sin (5 \times 1 + 1) d x

\int x^{3} \cdot ln (x^{2}) d x

\int e^{- s t} (t^{2} e^{3 t}) d t

\int \frac{x ^{2}}{( x - 8 ) ( x ^{2} - 64 )} d x