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人気のある微分積分 >

integral of (3x^2+1)^3x

解

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

解

\frac{1}{24} (3 x^{2} + 1)^{4} + C

解答ステップ

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

u置換積分法を適用する

= \int \frac{u ^{3}}{6} d u

定数を除く:

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

= \frac{1}{6} \cdot \int u^{3} d u

べき乗則を適用する

= \frac{1}{6} \cdot \frac{u ^{4}}{4}

代用を戻す

u = 3 x^{2} + 1

= \frac{1}{6} \cdot \frac{( 3 x ^{2} + 1 ) ^{4}}{4}

簡素化

\frac{1}{6} \cdot \frac{( 3 x ^{2} + 1 ) ^{4}}{4} : \frac{1}{24} (3 x^{2} + 1)^{4}

= \frac{1}{24} (3 x^{2} + 1)^{4}

定数を解答に追加する

= \frac{1}{24} (3 x^{2} + 1)^{4} + C

グラフ

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