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受欢迎的微积分 >

integral of (x^4)/((4-x^2)^{7/2)}

解答

\int \frac{x ^{4}}{( 4 - x ^{2} ) ^{\frac{7}{2}}} d x

解答

\frac{x ^{5}}{20 ( 4 - x ^{2} ) ^{\frac{5}{2}}} + C

求解步骤

\int \frac{x ^{4}}{( 4 - x ^{2} ) ^{\frac{7}{2}}} d x

使用三角换元法

= \int \frac{sin ^{4} ( u )}{4 cos ^{6} ( u )} d u

提出常数:

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

= \frac{1}{4} \cdot \int \frac{sin ^{4} ( u )}{cos ^{6} ( u )} d u

使用三角恒等式改写

= \frac{1}{4} \cdot \int \frac{tan ^{4} ( u )}{cos ^{2} ( u )} d u

使用基本三角恒等式:

\frac{1}{cos ( x )} = sec (x)

= \frac{1}{4} \cdot \int tan^{4} (u) sec^{2} (u) d u

使用三角恒等式改写

= \frac{1}{4} \cdot \int (1 + tan^{2} (u)) tan^{4} (u) d u

使用换元积分法

= \frac{1}{4} \cdot \int v^{4} d v

使用幂法则

= \frac{1}{4} \cdot \frac{v ^{5}}{5}

代回

= \frac{1}{4} \cdot \frac{tan ^{5} ( arcsin ( \frac{1}{2} x ) )}{5}

化简

\frac{1}{4} \cdot \frac{tan ^{5} ( arcsin ( \frac{1}{2} x ) )}{5} : \frac{x ^{5}}{20 ( 4 - x ^{2} ) ^{\frac{5}{2}}}

= \frac{x ^{5}}{20 ( 4 - x ^{2} ) ^{\frac{5}{2}}}

解答补常数

= \frac{x ^{5}}{20 ( 4 - x ^{2} ) ^{\frac{5}{2}}} + C

作图

流行的例子

integral of ((4x-5))/((x^2-2x+2)^{3/2)}

\int \frac{( 4 x - 5 )}{( x ^{2} - 2 x + 2 ) ^{\frac{3}{2}}} d x

integral of (5x-11)/(x^2+10x+9)

\int \frac{5 x - 11}{x ^{2} + 10 x + 9} d x

integral of 1/((x^2+49)^3)

\int \frac{1}{( x ^{2} + 49 ) ^{3}} d x

integral of 1/(u(u+1))

\int \frac{1}{u ( u + 1 )} d u

integral of 1/(sqrt(3x^2-5))

\int \frac{1}{3 x ^{2} - 5} d x