integral from 0 to y of sqrt(9+y^2)

解

\int_{0}^{y} 9 + y^{2} d y

9 (\frac{y y ^{2} + 9}{18} + \frac{1}{2} ln (\frac{1}{3} y + y^{2} + 9))

解答ステップ

\int_{0}^{y} 9 + y^{2} d y

三角関数による置換を適用する

= \int_{0}^{a r c t a n (\frac{y}{3})} 9 sec^{3} (u) d u

定数を除く:

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

= 9 \cdot \int_{0}^{a r c t a n (\frac{y}{3})} sec^{3} (u) d u

積分換算を適用する

= 9 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{y}{3})} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{y}{3})} sec (u) d u

\int_{0}^{a r c t a n (\frac{y}{3})} sec (u) d u = ln (\frac{1}{3} y + y^{2} + 9)

= 9 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{y}{3})} + \frac{1}{2} ln (\frac{1}{3} y + y^{2} + 9)

簡素化

9 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{y}{3})} + \frac{1}{2} ln (\frac{1}{3} y + y^{2} + 9) : 9 ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{y}{3})} + \frac{1}{2} ln (\frac{1}{3} y + y^{2} + 9))

= 9 ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{y}{3})} + \frac{1}{2} ln (\frac{1}{3} y + y^{2} + 9))

限度を計算する:

\frac{1}{18} y y^{2} + 9

= 9 (\frac{1}{18} y y^{2} + 9 + \frac{1}{2} ln (\frac{1}{3} y + y^{2} + 9))

簡素化

= 9 (\frac{y y ^{2} + 9}{18} + \frac{1}{2} ln (\frac{1}{3} y + y^{2} + 9))