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人気のある微分積分 >

integral of sqrt(4x^2+4x+5)

解

\int 4 x^{2} + 4 x + 5 d x

解

\frac{1}{4} (2 x + 1) 4 x^{2} + 4 x + 5 + ln x + \frac{1}{2} + \frac{1}{2} 4 x^{2} + 4 x + 5 + C

解答ステップ

\int 4 x^{2} + 4 x + 5 d x

平方完成する

4 x^{2} + 4 x + 5 : 4 (x + \frac{1}{2})^{2} + 4

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

u置換積分法を適用する

= \int 2 u^{2} + 1 d u

定数を除く:

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

= 2 \cdot \int u^{2} + 1 d u

三角関数による置換を適用する

= 2 \cdot \int sec^{3} (v) d v

積分換算を適用する

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

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

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

代用を戻す

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

簡素化

2 (\frac{sec ^{2} ( arctan ( x + \frac{1}{2} ) ) sin ( arctan ( x + \frac{1}{2} ) )}{2} + \frac{1}{2} ln tan (arctan (x + \frac{1}{2})) + sec (arctan (x + \frac{1}{2}))) : \frac{1}{4} (2 x + 1) 4 x^{2} + 4 x + 5 + ln x + \frac{1}{2} + \frac{1}{2} 4 x^{2} + 4 x + 5

= \frac{1}{4} (2 x + 1) 4 x^{2} + 4 x + 5 + ln x + \frac{1}{2} + \frac{1}{2} 4 x^{2} + 4 x + 5

定数を解答に追加する

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

グラフ

人気の例

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