integral of xln(x^2+1)+(x^2+1)y

解

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

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

解答ステップ

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

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

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

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

定数を解答に追加する

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