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人気のある微分積分 >

integral of sqrt(x^2+9^2)

解

\int x^{2} + 9^{2} d x

解

81 (\frac{1}{162} x 81 + x^{2} + \frac{1}{2} ln (\frac{x + 81 + x ^{2}}{9})) + C

解答ステップ

\int x^{2} + 9^{2} d x

簡素化

= \int x^{2} + 81 d x

三角関数による置換を適用する

= \int 81 sec^{3} (u) d u

定数を除く:

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

= 81 \cdot \int sec^{3} (u) d u

積分換算を適用する

= 81 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u)

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= 81 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

代用を戻す

u = arctan (\frac{1}{9} x)

= 81 (\frac{sec ^{2} ( arctan ( \frac{1}{9} x ) ) sin ( arctan ( \frac{1}{9} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{9} x)) + sec (arctan (\frac{1}{9} x)))

簡素化

81 (\frac{sec ^{2} ( arctan ( \frac{1}{9} x ) ) sin ( arctan ( \frac{1}{9} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{9} x)) + sec (arctan (\frac{1}{9} x))) : 81 (\frac{1}{162} x 81 + x^{2} + \frac{1}{2} ln (\frac{x + 81 + x ^{2}}{9}))

= 81 (\frac{1}{162} x 81 + x^{2} + \frac{1}{2} ln (\frac{x + 81 + x ^{2}}{9}))

定数を解答に追加する

= 81 (\frac{1}{162} x 81 + x^{2} + \frac{1}{2} ln (\frac{x + 81 + x ^{2}}{9})) + C

グラフ

人気の例

integral of xsqrt(9x-10)

\int x 9 x - 10 d x

integral of cos(7x)cos(6x)

\int cos (7 x) cos (6 x) d x

integral of (csc^6(2x))/(sqrt(cot(2x)))

\int \frac{csc ^{6} ( 2 x )}{cot ( 2 x )} d x

integral of (x^2+4)/(x(x-1)^2)

\int \frac{x ^{2} + 4}{x ( x - 1 ) ^{2}} d x

integral of 1/2 tcos(t^2)

\int \frac{1}{2} t cos (t^{2}) d t