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

解答

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

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

求解步骤

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

使用三角换元法

= \int cos (u) d u

使用积分常见定则:

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

= sin (u)

u = arctan (x)

代回

= sin (arctan (x))

利用以下特性:

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

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

解答补常数

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

\int 4 x^{3} + \frac{7}{x} d x

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

\int \frac{3}{2 x + 1} d x

\int \frac{1}{4 6 - 9 x} d x

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