integral of 1/((9-16x^2)^{3/2)}

解答

\int \frac{1}{( 9 - 16 x ^{2} ) ^{\frac{3}{2}}} d x

\frac{x}{9 ( - 16 x ^{2} + 9 ) ^{\frac{1}{2}}} + C

求解步骤

\int \frac{1}{( 9 - 16 x ^{2} ) ^{\frac{3}{2}}} d x

使用三角换元法

= \int \frac{1}{36 cos ^{2} ( u )} d u

提出常数:

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

= \frac{1}{36} \cdot \int \frac{1}{cos ^{2} ( u )} d u

使用积分常见定则:

\int \frac{1}{cos ^{2} ( u )} d u = tan (u)

= \frac{1}{36} tan (u)

u = arcsin (\frac{4}{3} x)

代回

= \frac{1}{36} tan (arcsin (\frac{4}{3} x))

化简

\frac{1}{36} tan (arcsin (\frac{4}{3} x)) : \frac{x}{9 ( - 16 x ^{2} + 9 ) ^{\frac{1}{2}}}

= \frac{x}{9 ( - 16 x ^{2} + 9 ) ^{\frac{1}{2}}}

解答补常数

= \frac{x}{9 ( - 16 x ^{2} + 9 ) ^{\frac{1}{2}}} + C

\int \frac{x ^{2}}{x ^{2} + 2 x + 5} d x

\int 2 cos^{3} (x) sin^{5} (x) d x

\int e^{e^{r} + r} cos (e^{r}) d r

\int \frac{x}{x ^{2} - 6 x + 13} d x

\int e^{t} (1 - e^{t})^{2} d t