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人気のある微分積分 >

integral of sqrt(x^2+5x)

解

\int x^{2} + 5 x d x

解

\frac{1}{8} 2 (2 x + 5) x (x + 5) - 25 ln \frac{2 x + 2 x ( x + 5 ) + 5}{5} + C

解答ステップ

\int x^{2} + 5 x d x

平方完成する

x^{2} + 5 x : (x + \frac{5}{2})^{2} - \frac{25}{4}

= \int (x + \frac{5}{2})^{2} - \frac{25}{4} d x

u置換積分法を適用する

= \int \frac{4 u ^{2} - 25}{2} d u

定数を除く:

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

= \frac{1}{2} \cdot \int 4 u^{2} - 25 d u

三角関数による置換を適用する

= \frac{1}{2} \cdot \int \frac{25}{2} 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

= \frac{1}{2} \cdot \frac{25}{2} \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

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

拡張

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

= \frac{1}{2} \cdot \frac{25}{2} \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

= \frac{1}{2} \cdot \frac{25}{2} (- \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) ∣

= \frac{1}{2} \cdot \frac{25}{2} (- 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

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

簡素化

\frac{1}{2} \cdot \frac{25}{2} (- ln tan (arcsec (\frac{2}{5} (x + \frac{5}{2}))) + sec (arcsec (\frac{2}{5} (x + \frac{5}{2}))) + \frac{1}{2} sec (arcsec (\frac{2}{5} (x + \frac{5}{2}))) tan (arcsec (\frac{2}{5} (x + \frac{5}{2}))) + \frac{1}{2} ln tan (arcsec (\frac{2}{5} (x + \frac{5}{2}))) + sec (arcsec (\frac{2}{5} (x + \frac{5}{2})))) : \frac{1}{8} 2 (2 x + 5) x (x + 5) - 25 ln \frac{2 x + 2 x ( x + 5 ) + 5}{5}

= \frac{1}{8} 2 (2 x + 5) x (x + 5) - 25 ln \frac{2 x + 2 x ( x + 5 ) + 5}{5}

定数を解答に追加する

= \frac{1}{8} 2 (2 x + 5) x (x + 5) - 25 ln \frac{2 x + 2 x ( x + 5 ) + 5}{5} + C

グラフ

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