integral from 4 to x of sqrt(x^2+9)

解

\int_{4}^{x} x^{2} + 9 d x

\frac{x x ^{2} + 9 + ln ( \frac{∣ x + 9 + x ^{2} ∣ ^{9}}{387420489} )}{2} - 10

解答ステップ

\int_{4}^{x} x^{2} + 9 d x

三角関数による置換を適用する

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

積分換算を適用する

= 9 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (\frac{4}{3})}^{a r c t a n (\frac{x}{3})} + \frac{1}{2} \cdot \int_{a r c t a n (\frac{4}{3})}^{a r c t a n (\frac{x}{3})} sec (u) d u

\int_{a r c t a n (\frac{4}{3})}^{a r c t a n (\frac{x}{3})} sec (u) d u = ln (\frac{1}{3} x + 9 + x^{2}) - ln (3)

= 9 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (\frac{4}{3})}^{a r c t a n (\frac{x}{3})} + \frac{1}{2} (ln (\frac{1}{3} x + 9 + x^{2}) - ln (3))

簡素化

9 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (\frac{4}{3})}^{a r c t a n (\frac{x}{3})} + \frac{1}{2} (ln (\frac{1}{3} x + 9 + x^{2}) - ln (3)) : 9 ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (\frac{4}{3})}^{a r c t a n (\frac{x}{3})} + \frac{1}{2} (ln (\frac{1}{3} x + 9 + x^{2}) - ln (3)))

= 9 ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (\frac{4}{3})}^{a r c t a n (\frac{x}{3})} + \frac{1}{2} (ln (\frac{1}{3} x + 9 + x^{2}) - ln (3)))

限度を計算する:

\frac{1}{18} x 9 + x^{2} - \frac{10}{9}

= 9 (\frac{1}{18} x 9 + x^{2} - \frac{10}{9} + \frac{1}{2} (ln (\frac{1}{3} x + 9 + x^{2}) - ln (3)))

簡素化

= \frac{x x ^{2} + 9 + ln ( \frac{∣ x + 9 + x ^{2} ∣ ^{9}}{387420489} )}{2} - 10