integral of (sqrt(64x^2-1))/x

解

\int \frac{64 x ^{2} - 1}{x} d x

- arctan (64 x^{2} - 1) + 64 x^{2} - 1 + C

解答ステップ

\int \frac{64 x ^{2} - 1}{x} d x

u置換積分法を適用する

= \int \frac{u ^{2}}{u ^{2} + 1} d u

\frac{u ^{2}}{u ^{2} + 1} = - \frac{1}{u ^{2} + 1} + 1

= \int - \frac{1}{u ^{2} + 1} + 1 d u

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{1}{u ^{2} + 1} d u + \int 1 d u

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

\int 1 d u = u

= - arctan (u) + u

代用を戻す

u = 64 x^{2} - 1

= - arctan (64 x^{2} - 1) + 64 x^{2} - 1

定数を解答に追加する

= - arctan (64 x^{2} - 1) + 64 x^{2} - 1 + C