integral of xsqrt(1+x)+(x^2+4)/(x-1)

解

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

\frac{2}{5} (1 + x)^{\frac{5}{2}} - \frac{2}{3} (1 + x)^{\frac{3}{2}} + \frac{( x - 1 ) ^{2}}{2} + 2 (x - 1) + 5 ln ∣ x - 1 ∣ + C

解答ステップ

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

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

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

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

定数を解答に追加する

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