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

解

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

x ln^{2} (x + 1 + x^{2}) - 2 (ln (x + 1 + x^{2}) 1 + x^{2} - x) + C

解答ステップ

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

部分積分を適用する

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

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

= x ln^{2} (x + 1 + x^{2}) - 2 (ln (x + 1 + x^{2}) 1 + x^{2} - x)

定数を解答に追加する

= x ln^{2} (x + 1 + x^{2}) - 2 (ln (x + 1 + x^{2}) 1 + x^{2} - x) + C