integral from 0 to x of sqrt(x^4+x^6)

解

\int_{0}^{x} x^{4} + x^{6} d x

- \frac{x x ^{2} + 1 + ln x + x ^{2} + 1}{2} + \frac{x ( x ^{2} + 1 ) ^{\frac{3}{2}}}{4} + \frac{3 ( x x ^{2} + 1 + ln x + x ^{2} + 1 )}{8}

解答ステップ

\int_{0}^{x} x^{4} + x^{6} d x

簡素化

x^{4} + x^{6} : x^{2} 1 + x^{2}

= \int_{0}^{x} x^{2} 1 + x^{2} d x

三角関数による置換を適用する

= \int_{0}^{a r c t a n (x)} sec^{3} (u) tan^{2} (u) d u

三角関数の公式を使用して書き換える

= \int_{0}^{a r c t a n (x)} sec^{3} (u) (- 1 + sec^{2} (u)) d u

拡張

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

= \int_{0}^{a r c t a n (x)} - sec^{3} (u) + sec^{5} (u) d u

総和規則を適用する:

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

= - \int_{0}^{a r c t a n (x)} sec^{3} (u) d u + \int_{0}^{a r c t a n (x)} sec^{5} (u) d u

\int_{0}^{a r c t a n (x)} sec^{3} (u) d u = \frac{x x ^{2} + 1 + ln x + x ^{2} + 1}{2}

\int_{0}^{a r c t a n (x)} sec^{5} (u) d u = \frac{x ( x ^{2} + 1 ) ^{\frac{3}{2}}}{4} + \frac{3 ( x x ^{2} + 1 + ln x + x ^{2} + 1 )}{8}

= - \frac{x x ^{2} + 1 + ln x + x ^{2} + 1}{2} + \frac{x ( x ^{2} + 1 ) ^{\frac{3}{2}}}{4} + \frac{3 ( x x ^{2} + 1 + ln x + x ^{2} + 1 )}{8}