integral from-1 to 1 of 1/((x^2+1)^2)

解答

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

\frac{π + 2}{4}

十进制

1.28539 \dots

求解步骤

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

使用三角换元法

= \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{1}{sec ^{4} ( u ) cos ^{2} ( u )} d u

使用三角恒等式改写

= \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{1 + cos ( 2 u )}{2} d u

提出常数:

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

= \frac{1}{2} \cdot \int_{- \frac{π}{4}}^{\frac{π}{4}} 1 + cos (2 u) d u

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{- \frac{π}{4}}^{\frac{π}{4}} 1 d u + \int_{- \frac{π}{4}}^{\frac{π}{4}} cos (2 u) d u)

\int_{- \frac{π}{4}}^{\frac{π}{4}} 1 d u = \frac{π}{2}

\int_{- \frac{π}{4}}^{\frac{π}{4}} cos (2 u) d u = 1

= \frac{1}{2} (\frac{π}{2} + 1)

化简

\frac{1}{2} (\frac{π}{2} + 1) : \frac{π + 2}{4}

= \frac{π + 2}{4}

\int_{1}^{4} \frac{- 3}{( 2 x + 1 ) ^{2}} d x

\int_{0}^{x} t^{6} 1 + t^{7} d t

\int_{4}^{1} 4 x + 3 x^{2} - 2 x^{3} d x

\int_{- a}^{a} (a^{2} - t^{2}) d t

\int_{0}^{4} (2 t + 1) d t