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

解

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

ln ∣ x ∣ + \frac{4}{5} x^{\frac{5}{2}} + \frac{x ^{5}}{5} + C

解答ステップ

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

拡張

(x^{2} + \frac{1}{x})^{2} : \frac{1}{x} + 2 x^{\frac{3}{2}} + x^{4}

= \int \frac{1}{x} + 2 x^{\frac{3}{2}} + x^{4} 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

= \int \frac{1}{x} d x + \int 2 x^{\frac{3}{2}} d x + \int x^{4} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int 2 x^{\frac{3}{2}} d x = \frac{4}{5} x^{\frac{5}{2}}

\int x^{4} d x = \frac{x ^{5}}{5}

= ln ∣ x ∣ + \frac{4}{5} x^{\frac{5}{2}} + \frac{x ^{5}}{5}

定数を解答に追加する

= ln ∣ x ∣ + \frac{4}{5} x^{\frac{5}{2}} + \frac{x ^{5}}{5} + C