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人気のある微分積分 >

integral of (x^2)/(16+9x^2)

解

\int \frac{x ^{2}}{16 + 9 x ^{2}} d x

解

\frac{1}{27} (3 x - 4 arctan (\frac{3}{4} x)) + C

解答ステップ

\int \frac{x ^{2}}{16 + 9 x ^{2}} d x

16 + 9 x^{2} = 16 (1 + \frac{9 x ^{2}}{16})

= \int \frac{x ^{2}}{16 ( 1 + \frac{9 x ^{2}}{16} )} d x

定数を除く:

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

= \frac{1}{16} \cdot \int \frac{x ^{2}}{1 + \frac{9 x ^{2}}{16}} d x

置換積分法を適用する

= \frac{1}{16} \cdot \int \frac{64 u ^{2}}{27 ( 1 + u ^{2} )} d u

定数を除く:

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

= \frac{1}{16} \cdot \frac{64}{27} \cdot \int \frac{u ^{2}}{1 + u ^{2}} d u

\frac{u ^{2}}{1 + u ^{2}} = - \frac{1}{1 + u ^{2}} + 1

= \frac{1}{16} \cdot \frac{64}{27} \cdot \int - \frac{1}{1 + u ^{2}} + 1 d u

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}{16} \cdot \frac{64}{27} (- \int \frac{1}{1 + u ^{2}} d u + \int 1 d u)

\int \frac{1}{1 + u ^{2}} d u = arctan (u)

\int 1 d u = u

= \frac{1}{16} \cdot \frac{64}{27} (- arctan (u) + u)

代用を戻す

u = \frac{3}{4} x

= \frac{1}{16} \cdot \frac{64}{27} (- arctan (\frac{3}{4} x) + \frac{3}{4} x)

簡素化

\frac{1}{16} \cdot \frac{64}{27} (- arctan (\frac{3}{4} x) + \frac{3}{4} x) : \frac{1}{27} (3 x - 4 arctan (\frac{3}{4} x))

= \frac{1}{27} (3 x - 4 arctan (\frac{3}{4} x))

定数を解答に追加する

= \frac{1}{27} (3 x - 4 arctan (\frac{3}{4} x)) + C

グラフ

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