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

解

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

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

解答ステップ

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

三角関数による置換を適用する

= \int_{a r c t a n (x)}^{\frac{π}{4}} sec^{3} (u) d u

積分換算を適用する

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (x)}^{\frac{π}{4}} + \frac{1}{2} \cdot \int_{a r c t a n (x)}^{\frac{π}{4}} sec (u) d u

\int_{a r c t a n (x)}^{\frac{π}{4}} sec (u) d u = ln (1 + 2) - ln x + 1 + x^{2}

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (x)}^{\frac{π}{4}} + \frac{1}{2} (ln (1 + 2) - ln x + 1 + x^{2})

簡素化

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (x)}^{\frac{π}{4}} + \frac{1}{2} (ln (1 + 2) - ln x + 1 + x^{2}) : [\frac{1}{2} sec (u) tan (u)]_{a r c t a n (x)}^{\frac{π}{4}} + \frac{1}{2} (ln (1 + 2) - ln x + 1 + x^{2})

= [\frac{1}{2} sec (u) tan (u)]_{a r c t a n (x)}^{\frac{π}{4}} + \frac{1}{2} (ln (1 + 2) - ln x + 1 + x^{2})

限度を計算する:

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

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

簡素化

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