integral of ((x^2)/4+4x+19)^2

解

\int (\frac{x ^{2}}{4} + 4 x + 19)^{2} d x

\frac{x ^{5}}{80} + \frac{x ^{4}}{2} + \frac{17 x ^{3}}{2} + 76 x^{2} + 361 x + C

解答ステップ

\int (\frac{x ^{2}}{4} + 4 x + 19)^{2} d x

拡張

(\frac{x ^{2}}{4} + 4 x + 19)^{2} : \frac{x ^{4}}{16} + 2 x^{3} + \frac{51 x ^{2}}{2} + 152 x + 361

= \int \frac{x ^{4}}{16} + 2 x^{3} + \frac{51 x ^{2}}{2} + 152 x + 361 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{x ^{4}}{16} d x + \int 2 x^{3} d x + \int \frac{51 x ^{2}}{2} d x + \int 152 x d x + \int 361 d x

\int \frac{x ^{4}}{16} d x = \frac{x ^{5}}{80}

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

\int \frac{51 x ^{2}}{2} d x = \frac{17 x ^{3}}{2}

\int 152 x d x = 76 x^{2}

\int 361 d x = 361 x

= \frac{x ^{5}}{80} + \frac{x ^{4}}{2} + \frac{17 x ^{3}}{2} + 76 x^{2} + 361 x

定数を解答に追加する

= \frac{x ^{5}}{80} + \frac{x ^{4}}{2} + \frac{17 x ^{3}}{2} + 76 x^{2} + 361 x + C