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人気のある微分積分 >

integral from 1 to 8 of 1/(xsqrt(16x^2-9))

解

\int_{1}^{8} \frac{1}{x 16 x ^{2} - 9} d x

解

\frac{1}{3} (arcsec (\frac{32}{3}) - arcsec (\frac{4}{3}))

+1

十進法表記

0.25139 \dots

解答ステップ

\int_{1}^{8} \frac{1}{x 16 x ^{2} - 9} d x

三角関数による置換を適用する

= \int_{arcsec (\frac{4}{3})}^{arcsec (\frac{32}{3})} \frac{1}{3} d u

定数を除く:

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

= \frac{1}{3} \cdot \int_{arcsec (\frac{4}{3})}^{arcsec (\frac{32}{3})} 1 d u

定数の積分:

\int a d x = a x

= \frac{1}{3} [1 \cdot u]_{arcsec (\frac{4}{3})}^{arcsec (\frac{32}{3})}

簡素化

= \frac{1}{3} [u]_{arcsec (\frac{4}{3})}^{arcsec (\frac{32}{3})}

限度を計算する:

arcsec (\frac{32}{3}) - arcsec (\frac{4}{3})

= \frac{1}{3} (arcsec (\frac{32}{3}) - arcsec (\frac{4}{3}))

グラフ

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