integral from 5 to x^2 of 4sqrt(1+x^2)

解

\int_{5}^{x^{2}} 4 1 + x^{2} d x

4 (\frac{x ^{2} 1 + x ^{4}}{2} - \frac{5 13}{2} + \frac{1}{2} (ln x^{2} + 1 + x^{4} - ln (5 + 26)))

解答ステップ

\int_{5}^{x^{2}} 4 1 + x^{2} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int_{5}^{x^{2}} 1 + x^{2} d x

三角関数による置換を適用する

= 4 \cdot \int_{a r c t a n (5)}^{a r c t a n (x^{2})} sec^{3} (u) d u

積分換算を適用する

= 4 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (5)}^{a r c t a n (x^{2})} + \frac{1}{2} \cdot \int_{a r c t a n (5)}^{a r c t a n (x^{2})} sec (u) d u

\int_{a r c t a n (5)}^{a r c t a n (x^{2})} sec (u) d u = ln x^{2} + 1 + x^{4} - ln (5 + 26)

= 4 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (5)}^{a r c t a n (x^{2})} + \frac{1}{2} (ln x^{2} + 1 + x^{4} - ln (5 + 26))

簡素化

4 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (5)}^{a r c t a n (x^{2})} + \frac{1}{2} (ln x^{2} + 1 + x^{4} - ln (5 + 26)) : 4 ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (5)}^{a r c t a n (x^{2})} + \frac{1}{2} (ln x^{2} + 1 + x^{4} - ln (5 + 26)))

= 4 ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (5)}^{a r c t a n (x^{2})} + \frac{1}{2} (ln x^{2} + 1 + x^{4} - ln (5 + 26)))

限度を計算する:

\frac{1}{2} x^{2} 1 + x^{4} - \frac{5 13}{2}

= 4 (\frac{1}{2} x^{2} 1 + x^{4} - \frac{5 13}{2} + \frac{1}{2} (ln x^{2} + 1 + x^{4} - ln (5 + 26)))

簡素化

= 4 (\frac{x ^{2} 1 + x ^{4}}{2} - \frac{5 13}{2} + \frac{1}{2} (ln x^{2} + 1 + x^{4} - ln (5 + 26)))