integral from 0 to x of sqrt(1+t^2)

解

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

\frac{x x ^{2} + 1 + ln x + x ^{2} + 1}{2}

解答ステップ

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

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) d u

積分換算を適用する

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (x)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (x)} sec (u) d u

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

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (x)} + \frac{1}{2} ln x + x^{2} + 1

簡素化

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (x)} + \frac{1}{2} ln x + x^{2} + 1 : [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (x)} + \frac{1}{2} ln x + x^{2} + 1

= [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (x)} + \frac{1}{2} ln x + x^{2} + 1

限度を計算する:

\frac{1}{2} x x^{2} + 1

= \frac{1}{2} x x^{2} + 1 + \frac{1}{2} ln x + x^{2} + 1

簡素化

= \frac{x x ^{2} + 1 + ln x + x ^{2} + 1}{2}