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人気のある微分積分 >

integral of sqrt(x^2+6x+5)

解

\int x^{2} + 6 x + 5 d x

解

4 (- ln (\frac{1}{2} x^{2} + 6 x + 5 + x + 3) + \frac{1}{8} (x + 3) x^{2} + 6 x + 5 + \frac{1}{2} ln (\frac{x ^{2} + 6 x + 5 + x + 3}{2})) + C

解答ステップ

\int x^{2} + 6 x + 5 d x

平方完成する

x^{2} + 6 x + 5 : (x + 3)^{2} - 4

= \int (x + 3)^{2} - 4 d x

u置換積分法を適用する

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

三角関数による置換を適用する

= \int 4 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

= 4 \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

= 4 \cdot \int (- 1 + sec^{2} (v)) sec (v) d v

拡張

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

= 4 \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

= 4 (- \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) ∣

= 4 (- 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

= 4 (- ln tan (arcsec (\frac{1}{2} (x + 3))) + sec (arcsec (\frac{1}{2} (x + 3))) + \frac{1}{2} sec (arcsec (\frac{1}{2} (x + 3))) tan (arcsec (\frac{1}{2} (x + 3))) + \frac{1}{2} ln tan (arcsec (\frac{1}{2} (x + 3))) + sec (arcsec (\frac{1}{2} (x + 3))))

簡素化

4 (- ln tan (arcsec (\frac{1}{2} (x + 3))) + sec (arcsec (\frac{1}{2} (x + 3))) + \frac{1}{2} sec (arcsec (\frac{1}{2} (x + 3))) tan (arcsec (\frac{1}{2} (x + 3))) + \frac{1}{2} ln tan (arcsec (\frac{1}{2} (x + 3))) + sec (arcsec (\frac{1}{2} (x + 3)))) : 4 (- ln (\frac{1}{2} x^{2} + 6 x + 5 + x + 3) + \frac{1}{8} (x + 3) x^{2} + 6 x + 5 + \frac{1}{2} ln (\frac{x ^{2} + 6 x + 5 + x + 3}{2}))

= 4 (- ln (\frac{1}{2} x^{2} + 6 x + 5 + x + 3) + \frac{1}{8} (x + 3) x^{2} + 6 x + 5 + \frac{1}{2} ln (\frac{x ^{2} + 6 x + 5 + x + 3}{2}))

定数を解答に追加する

= 4 (- ln (\frac{1}{2} x^{2} + 6 x + 5 + x + 3) + \frac{1}{8} (x + 3) x^{2} + 6 x + 5 + \frac{1}{2} ln (\frac{x ^{2} + 6 x + 5 + x + 3}{2})) + C

グラフ

人気の例

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