integral of+(13)/((t^2+1)^{3/2)}

解答

\int + \frac{13}{( t ^{2} + 1 ) ^{\frac{3}{2}}} d t

\frac{13 t}{( 1 + t ^{2} ) ^{\frac{1}{2}}} + C

求解步骤

\int \frac{13}{( t ^{2} + 1 ) ^{\frac{3}{2}}} d t

提出常数:

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

= 13 \cdot \int \frac{1}{( t ^{2} + 1 ) ^{\frac{3}{2}}} d t

使用三角换元法

= 13 \cdot \int cos (u) d u

使用积分常见定则:

\int cos (u) d u = sin (u)

= 13 sin (u)

u = arctan (t)

代回

= 13 sin (arctan (t))

化简

13 sin (arctan (t)) : \frac{13 t}{( 1 + t ^{2} ) ^{\frac{1}{2}}}

= \frac{13 t}{( 1 + t ^{2} ) ^{\frac{1}{2}}}

解答补常数

= \frac{13 t}{( 1 + t ^{2} ) ^{\frac{1}{2}}} + C

x \to 1 lim (\frac{1}{x})

n \to \infty lim (tanh (n))

\int \frac{x ^{2} + x + 1}{( x + 2 ) ( x + 1 ) ^{2}} d x

\int 4 e^{- t} t d t

\int sec (8 x) d x