integral from 1 to x of ln(7+t^2)

解

\int_{1}^{x} ln (7 + t^{2}) d t

x ln (7 + x^{2}) + 27 arctan (\frac{x}{7}) - 2 x - 3 ln (2) - 27 arctan (\frac{1}{7}) + 2

解答ステップ

\int_{1}^{x} ln (7 + t^{2}) d t

部分積分を適用する

= [t ln (7 + t^{2}) - \int \frac{2 t ^{2}}{7 + t ^{2}} d t]_{1}^{x}

\int \frac{2 t ^{2}}{7 + t ^{2}} d t = - 27 arctan (\frac{t}{7}) + 2 t

= [t ln (7 + t^{2}) - (- 27 arctan (\frac{t}{7}) + 2 t)]_{1}^{x}

簡素化

= [t ln (7 + t^{2}) + 27 arctan (\frac{t}{7}) - 2 t]_{1}^{x}

限度を計算する:

x ln (7 + x^{2}) + 27 arctan (\frac{x}{7}) - 2 x - 3 ln (2) - 27 arctan (\frac{1}{7}) + 2

= x ln (7 + x^{2}) + 27 arctan (\frac{x}{7}) - 2 x - 3 ln (2) - 27 arctan (\frac{1}{7}) + 2