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人気のある微分積分 >

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

解

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

解

\frac{5}{2} arctan (\frac{x}{2}) + x + C

解答ステップ

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

\frac{x ^{2} + 9}{x ^{2} + 4} = \frac{5}{x ^{2} + 4} + 1

= \int \frac{5}{x ^{2} + 4} + 1 d x

置換積分法を適用する

= \int 2 (\frac{5}{4 u ^{2} + 4} + 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 \frac{5}{4 u ^{2} + 4} + 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

= 2 (\int \frac{5}{4 u ^{2} + 4} d u + \int 1 d u)

\int \frac{5}{4 u ^{2} + 4} d u = \frac{5}{4} arctan (u)

\int 1 d u = u

= 2 (\frac{5}{4} arctan (u) + u)

代用を戻す

u = \frac{x}{2}

= 2 (\frac{5}{4} arctan (\frac{x}{2}) + \frac{x}{2})

簡素化

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

= \frac{5}{2} arctan (\frac{x}{2}) + x

定数を解答に追加する

= \frac{5}{2} arctan (\frac{x}{2}) + x + C

グラフ

人気の例

integral of 1/(7y)

\int \frac{1}{7 y} d y

(dy)/(dx)=1-y/x

\frac{d y}{d x} = 1 - \frac{y}{x}

derivative of (arctanh(sqrt(3x^2))^3)

\frac{d}{d x} ((arctanh (3 x^{2}))^{3})

integral of xsin(z)

\int x sin (z) d y

tangent f(x)=(x^3-16x)^8,\at x=4

t an g e n t f (x) = (x^{3} - 16 x)^{8}, at x = 4