integral of 6/(x(x^2+25))

解

\int \frac{6}{x ( x ^{2} + 25 )} d x

6 (\frac{1}{25} ln ∣ x ∣ - \frac{1}{50} ln x^{2} + 25) + C

解答ステップ

\int \frac{6}{x ( x ^{2} + 25 )} d x

定数を除く:

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

= 6 \cdot \int \frac{1}{x ( x ^{2} + 25 )} d x

以下の部分分数を得る:

\frac{1}{x ( x ^{2} + 25 )} : \frac{1}{25 x} - \frac{x}{25 ( x ^{2} + 25 )}

= 6 \cdot \int \frac{1}{25 x} - \frac{x}{25 ( x ^{2} + 25 )} d x

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

= 6 (\int \frac{1}{25 x} d x - \int \frac{x}{25 ( x ^{2} + 25 )} d x)

\int \frac{1}{25 x} d x = \frac{1}{25} ln ∣ x ∣

\int \frac{x}{25 ( x ^{2} + 25 )} d x = \frac{1}{50} ln x^{2} + 25

= 6 (\frac{1}{25} ln ∣ x ∣ - \frac{1}{50} ln x^{2} + 25)

定数を解答に追加する

= 6 (\frac{1}{25} ln ∣ x ∣ - \frac{1}{50} ln x^{2} + 25) + C