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人気のある微分積分 >

integral of 1/(sqrt(x^2+4/5))

解

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

解

ln (\frac{1}{2} 5 x + 4 + 5 x^{2}) + C

解答ステップ

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

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

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

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

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

定数を除く:

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

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

\frac{1}{\frac{4}{5}} = \frac{5}{2}

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

三角関数による置換を適用する

= \frac{5}{2} \cdot \int \frac{2 sec ( u )}{5} d u

定数を除く:

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

= \frac{5}{2} \cdot \frac{2}{5} \cdot \int sec (u) d u

共通積分を使用する:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= \frac{5}{2} \cdot \frac{2}{5} ln ∣ tan (u) + sec (u) ∣

代用を戻す

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

= \frac{5}{2} \cdot \frac{2}{5} ln tan (arctan (\frac{5}{2} x)) + sec (arctan (\frac{5}{2} x))

簡素化

\frac{5}{2} \cdot \frac{2}{5} ln tan (arctan (\frac{5}{2} x)) + sec (arctan (\frac{5}{2} x)) : ln (\frac{1}{2} 5 x + 4 + 5 x^{2})

= ln (\frac{1}{2} 5 x + 4 + 5 x^{2})

定数を解答に追加する

= ln (\frac{1}{2} 5 x + 4 + 5 x^{2}) + C

グラフ

人気の例

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