integral of (9-x^2)/(5x^3+x)

解

\int \frac{9 - x ^{2}}{5 x ^{3} + x} d x

9 ln ∣ x ∣ - \frac{23}{5} ln 5 x^{2} + 1 + C

解答ステップ

\int \frac{9 - x ^{2}}{5 x ^{3} + x} d x

拡張

\frac{9 - x ^{2}}{5 x ^{3} + x} : \frac{9}{5 x ^{3} + x} - \frac{x ^{2}}{5 x ^{3} + 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

= \int \frac{9}{5 x ^{3} + x} d x - \int \frac{x ^{2}}{5 x ^{3} + x} d x

\int \frac{9}{5 x ^{3} + x} d x = 9 (ln ∣ x ∣ - \frac{1}{2} ln 5 x^{2} + 1)

\int \frac{x ^{2}}{5 x ^{3} + x} d x = \frac{1}{10} ln 5 x^{2} + 1

= 9 (ln ∣ x ∣ - \frac{1}{2} ln 5 x^{2} + 1) - \frac{1}{10} ln 5 x^{2} + 1

簡素化

9 (ln ∣ x ∣ - \frac{1}{2} ln 5 x^{2} + 1) - \frac{1}{10} ln 5 x^{2} + 1 : 9 ln ∣ x ∣ - \frac{23}{5} ln 5 x^{2} + 1

= 9 ln ∣ x ∣ - \frac{23}{5} ln 5 x^{2} + 1

定数を解答に追加する

= 9 ln ∣ x ∣ - \frac{23}{5} ln 5 x^{2} + 1 + C