integral of 7/((x^2+1)^{3/2)}

解答

\int \frac{7}{( x ^{2} + 1 ) ^{\frac{3}{2}}} d x

\frac{7 x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + C

求解步骤

\int \frac{7}{( x ^{2} + 1 ) ^{\frac{3}{2}}} d x

提出常数:

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

= 7 \cdot \int \frac{1}{( x ^{2} + 1 ) ^{\frac{3}{2}}} d x

使用三角换元法

= 7 \cdot \int cos (u) d u

使用积分常见定则:

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

= 7 sin (u)

u = arctan (x)

代回

= 7 sin (arctan (x))

化简

7 sin (arctan (x)) : \frac{7 x}{( 1 + x ^{2} ) ^{\frac{1}{2}}}

= \frac{7 x}{( 1 + x ^{2} ) ^{\frac{1}{2}}}

解答补常数

= \frac{7 x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + C

\int_{0}^{4} \frac{x}{x ^{2} - 9} d x

\int \frac{5 x + 4}{x - 1} d x

\frac{d ^{2}}{d x ^{2}} (5 e^{x} cos (x))

x \frac{d y}{d x} + x y = 1 - y, y (1) = 0

(e^{x} + y) d x + (2 + x + y e^{y}) d y = 0, y (0) = 0