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人気のある微分積分 >

integral of sqrt(x^2+x+1)

解

\int x^{2} + x + 1 d x

解

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

解答ステップ

\int x^{2} + x + 1 d x

平方完成する

x^{2} + x + 1 : (x + \frac{1}{2})^{2} + \frac{3}{4}

= \int (x + \frac{1}{2})^{2} + \frac{3}{4} d x

u置換積分法を適用する

= \int \frac{4 u ^{2} + 3}{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} + 3 d u

三角関数による置換を適用する

= \frac{1}{2} \cdot \int \frac{3}{2} sec^{3} (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{3}{2} \cdot \int sec^{3} (v) d v

積分換算を適用する

= \frac{1}{2} \cdot \frac{3}{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) ∣

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

代用を戻す

= \frac{1}{2} \cdot \frac{3}{2} \frac{sec ^{2} ( arctan ( \frac{2}{3} ( x + \frac{1}{2} ) ) ) sin ( arctan ( \frac{2}{3} ( x + \frac{1}{2} ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{3} (x + \frac{1}{2}))) + sec (arctan (\frac{2}{3} (x + \frac{1}{2})))

簡素化

\frac{1}{2} \cdot \frac{3}{2} \frac{sec ^{2} ( arctan ( \frac{2}{3} ( x + \frac{1}{2} ) ) ) sin ( arctan ( \frac{2}{3} ( x + \frac{1}{2} ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{3} (x + \frac{1}{2}))) + sec (arctan (\frac{2}{3} (x + \frac{1}{2}))) : \frac{1}{8} (2 (2 x + 1) x^{2} + x + 1 + 3 (ln 2 x + 3 \frac{4 x ^{2} + 4 x + 4}{3} + 1 - \frac{ln ( 3 )}{2}))

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

定数を解答に追加する

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

グラフ

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