integral from x to-2 of ln(1+t^2)

解

\int_{x}^{- 2} ln (1 + t^{2}) d t

- 2 ln (5) - 2 arctan (2) + 4 - x ln (1 + x^{2}) - 2 arctan (x) + 2 x

解答ステップ

\int_{x}^{- 2} ln (1 + t^{2}) d t

部分積分を適用する

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

\int \frac{2 t ^{2}}{1 + t ^{2}} d t = 2 (- arctan (t) + t)

= [t ln (1 + t^{2}) - 2 (- arctan (t) + t)]_{x}^{- 2}

限度を計算する:

- 2 ln (5) - 2 arctan (2) + 4 - x ln (1 + x^{2}) - 2 arctan (x) + 2 x

= - 2 ln (5) - 2 arctan (2) + 4 - x ln (1 + x^{2}) - 2 arctan (x) + 2 x