integral from 7 to x of sqrt(t^2+15)

解

\int_{7}^{x} t^{2} + 15 d t

\frac{x x ^{2} + 15}{2} - 28 + \frac{15}{2} ln \frac{x}{15} + \frac{15 + x ^{2}}{15} - \frac{15}{4} ln (15)

解答ステップ

\int_{7}^{x} t^{2} + 15 d t

三角関数による置換を適用する

= \int_{a r c t a n (\frac{7}{15})}^{a r c t a n (\frac{x}{15})} 15 sec^{3} (u) d u

定数を除く:

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

= 15 \cdot \int_{a r c t a n (\frac{7}{15})}^{a r c t a n (\frac{x}{15})} sec^{3} (u) d u

積分換算を適用する

= 15 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (\frac{7}{15})}^{a r c t a n (\frac{x}{15})} + \frac{1}{2} \cdot \int_{a r c t a n (\frac{7}{15})}^{a r c t a n (\frac{x}{15})} sec (u) d u

\int_{a r c t a n (\frac{7}{15})}^{a r c t a n (\frac{x}{15})} sec (u) d u = ln \frac{x}{15} + \frac{15 + x ^{2}}{15} - \frac{1}{2} ln (15)

= 15 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (\frac{7}{15})}^{a r c t a n (\frac{x}{15})} + \frac{1}{2} (ln \frac{x}{15} + \frac{15 + x ^{2}}{15} - \frac{1}{2} ln (15))

簡素化

15 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (\frac{7}{15})}^{a r c t a n (\frac{x}{15})} + \frac{1}{2} (ln \frac{x}{15} + \frac{15 + x ^{2}}{15} - \frac{1}{2} ln (15)) : 15 [\frac{1}{2} sec (u) tan (u)]_{a r c t a n (\frac{7}{15})}^{a r c t a n (\frac{x}{15})} + \frac{15}{2} ln \frac{x}{15} + \frac{15 + x ^{2}}{15} - \frac{15}{4} ln (15)

= 15 [\frac{1}{2} sec (u) tan (u)]_{a r c t a n (\frac{7}{15})}^{a r c t a n (\frac{x}{15})} + \frac{15}{2} ln \frac{x}{15} + \frac{15 + x ^{2}}{15} - \frac{15}{4} ln (15)

限度を計算する:

\frac{x 15 + x ^{2}}{30} - \frac{28}{15}

= 15 (\frac{x 15 + x ^{2}}{30} - \frac{28}{15}) + \frac{15}{2} ln \frac{x}{15} + \frac{15 + x ^{2}}{15} - \frac{15}{4} ln (15)

簡素化

= \frac{x x ^{2} + 15}{2} - 28 + \frac{15}{2} ln \frac{x}{15} + \frac{15 + x ^{2}}{15} - \frac{15}{4} ln (15)