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人気のある微分積分 >

integral of (x^2)/(9x^2+4)

解

\int \frac{x ^{2}}{9 x ^{2} + 4} d x

解

\frac{1}{27} (3 x - 2 arctan (\frac{3}{2} x)) + C

解答ステップ

\int \frac{x ^{2}}{9 x ^{2} + 4} d x

9 x^{2} + 4 = 4 (\frac{9 x ^{2}}{4} + 1)

= \int \frac{x ^{2}}{4 ( \frac{9 x ^{2}}{4} + 1 )} d x

定数を除く:

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

= \frac{1}{4} \cdot \int \frac{x ^{2}}{\frac{9 x ^{2}}{4} + 1} d x

置換積分法を適用する

= \frac{1}{4} \cdot \int \frac{8 u ^{2}}{27 ( u ^{2} + 1 )} d u

定数を除く:

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

= \frac{1}{4} \cdot \frac{8}{27} \cdot \int \frac{u ^{2}}{u ^{2} + 1} d u

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

= \frac{1}{4} \cdot \frac{8}{27} \cdot \int - \frac{1}{u ^{2} + 1} + 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}{4} \cdot \frac{8}{27} (- \int \frac{1}{u ^{2} + 1} d u + \int 1 d u)

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

\int 1 d u = u

= \frac{1}{4} \cdot \frac{8}{27} (- arctan (u) + u)

代用を戻す

u = \frac{3}{2} x

= \frac{1}{4} \cdot \frac{8}{27} (- arctan (\frac{3}{2} x) + \frac{3}{2} x)

簡素化

\frac{1}{4} \cdot \frac{8}{27} (- arctan (\frac{3}{2} x) + \frac{3}{2} x) : \frac{1}{27} (3 x - 2 arctan (\frac{3}{2} x))

= \frac{1}{27} (3 x - 2 arctan (\frac{3}{2} x))

定数を解答に追加する

= \frac{1}{27} (3 x - 2 arctan (\frac{3}{2} x)) + C

グラフ

人気の例

integral of ((-1)/(x^2))

\int (\frac{- 1}{x ^{2}}) d x

integral of (10)/(x^3-x^2+9x-9)

\int \frac{10}{x ^{3} - x ^{2} + 9 x - 9} d x

integral of 1/(1-3x^2)

\int \frac{1}{1 - 3 x ^{2}} d x

integral of y/(sqrt(16-y^2))

\int \frac{y}{16 - y ^{2}} d y

integral of (7-3x)^5

\int (7 - 3 x)^{5} d x