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人気のある微分積分 >

integral of sqrt(x^2+10x)

解

\int x^{2} + 10 x d x

解

25 (- ln (\frac{1}{5} x^{2} + 10 x + x + 5) + \frac{1}{50} (x + 5) x^{2} + 10 x + \frac{1}{2} ln (\frac{x ^{2} + 10 x + x + 5}{5})) + C

解答ステップ

\int x^{2} + 10 x d x

平方完成する

x^{2} + 10 x : (x + 5)^{2} - 25

= \int (x + 5)^{2} - 25 d x

u置換積分法を適用する

= \int u^{2} - 25 d u

三角関数による置換を適用する

= \int 25 tan^{2} (v) sec (v) d v

定数を除く:

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

= 25 \cdot \int tan^{2} (v) sec (v) d v

三角関数の公式を使用して書き換える

= 25 \cdot \int (- 1 + sec^{2} (v)) sec (v) d v

拡張

(- 1 + sec^{2} (v)) sec (v) : - sec (v) + sec^{3} (v)

= 25 \cdot \int - sec (v) + sec^{3} (v) d v

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 25 (- \int sec (v) d v + \int sec^{3} (v) d v)

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

\int sec^{3} (v) d v = \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

= 25 (- ln ∣ tan (v) + sec (v) ∣ + \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

代用を戻す

= 25 (- ln tan (arcsec (\frac{1}{5} (x + 5))) + sec (arcsec (\frac{1}{5} (x + 5))) + \frac{1}{2} sec (arcsec (\frac{1}{5} (x + 5))) tan (arcsec (\frac{1}{5} (x + 5))) + \frac{1}{2} ln tan (arcsec (\frac{1}{5} (x + 5))) + sec (arcsec (\frac{1}{5} (x + 5))))

簡素化

25 (- ln tan (arcsec (\frac{1}{5} (x + 5))) + sec (arcsec (\frac{1}{5} (x + 5))) + \frac{1}{2} sec (arcsec (\frac{1}{5} (x + 5))) tan (arcsec (\frac{1}{5} (x + 5))) + \frac{1}{2} ln tan (arcsec (\frac{1}{5} (x + 5))) + sec (arcsec (\frac{1}{5} (x + 5)))) : 25 (- ln (\frac{1}{5} x^{2} + 10 x + x + 5) + \frac{1}{50} (x + 5) x^{2} + 10 x + \frac{1}{2} ln (\frac{x ^{2} + 10 x + x + 5}{5}))

= 25 (- ln (\frac{1}{5} x^{2} + 10 x + x + 5) + \frac{1}{50} (x + 5) x^{2} + 10 x + \frac{1}{2} ln (\frac{x ^{2} + 10 x + x + 5}{5}))

定数を解答に追加する

= 25 (- ln (\frac{1}{5} x^{2} + 10 x + x + 5) + \frac{1}{50} (x + 5) x^{2} + 10 x + \frac{1}{2} ln (\frac{x ^{2} + 10 x + x + 5}{5})) + C

グラフ

人気の例

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