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

解

\int_{x}^{1} x^{2} u^{2} + 1 d u

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

解答ステップ

\int_{x}^{1} x^{2} u^{2} + 1 d u

定数を除く:

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

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

三角関数による置換を適用する

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

積分換算を適用する

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

\int_{a r c t a n (x)}^{\frac{π}{4}} sec (v) d v = ln (1 + 2) - ln x + 1 + x^{2}

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

簡素化

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

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

限度を計算する:

\frac{1}{2} - \frac{1}{2} x 1 + x^{2}

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

簡素化

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