integral from-1 to 1 of 3x^2sqrt(x^2+1)

解答

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

3 (- \frac{1}{4} (2 + \frac{1}{2} (ln (1 + 2) - ln (- 1 + 2))) + 2)

十进制

2.52095 \dots

求解步骤

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

提出常数:

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

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

使用三角换元法

= 3 \cdot \int_{- \frac{π}{4}}^{\frac{π}{4}} sec^{3} (u) tan^{2} (u) d u

使用三角恒等式改写

= 3 \cdot \int_{- \frac{π}{4}}^{\frac{π}{4}} sec^{3} (u) (- 1 + sec^{2} (u)) d u

乘开

sec^{3} (u) (- 1 + sec^{2} (u)) : - sec^{3} (u) + sec^{5} (u)

= 3 \cdot \int_{- \frac{π}{4}}^{\frac{π}{4}} - sec^{3} (u) + sec^{5} (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

= 3 (- \int_{- \frac{π}{4}}^{\frac{π}{4}} sec^{3} (u) d u + \int_{- \frac{π}{4}}^{\frac{π}{4}} sec^{5} (u) d u)

\int_{- \frac{π}{4}}^{\frac{π}{4}} sec^{3} (u) d u = 2 + \frac{1}{2} (ln (1 + 2) - ln (- 1 + 2))

\int_{- \frac{π}{4}}^{\frac{π}{4}} sec^{5} (u) d u = 2 + \frac{3}{4} (2 + \frac{1}{2} (ln (1 + 2) - ln (- 1 + 2)))

= 3 (- (2 + \frac{1}{2} (ln (1 + 2) - ln (- 1 + 2))) + 2 + \frac{3}{4} (2 + \frac{1}{2} (ln (1 + 2) - ln (- 1 + 2))))

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