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人気のある微分積分 >

integral of (y^2)/((16y^2+9)^{3/2)}

解

\int \frac{y ^{2}}{( 16 y ^{2} + 9 ) ^{\frac{3}{2}}} d y

解

\frac{1}{64} (- \frac{4 y}{( 9 + 16 y ^{2} ) ^{\frac{1}{2}}} + ln (\frac{1}{3} 4 y + (9 + 16 y^{2})^{\frac{1}{2}})) + C

解答ステップ

\int \frac{y ^{2}}{( 16 y ^{2} + 9 ) ^{\frac{3}{2}}} d y

三角関数による置換を適用する

= \int \frac{tan ^{2} ( u )}{64 sec ( u )} d u

定数を除く:

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

= \frac{1}{64} \cdot \int \frac{tan ^{2} ( u )}{sec ( u )} d u

三角関数の公式を使用して書き換える

= \frac{1}{64} \cdot \int \frac{- 1 + sec ^{2} ( u )}{sec ( u )} d u

拡張

\frac{- 1 + sec ^{2} ( u )}{sec ( u )} : - \frac{1}{sec ( u )} + sec (u)

= \frac{1}{64} \cdot \int - \frac{1}{sec ( u )} + sec (u) 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

= \frac{1}{64} (- \int \frac{1}{sec ( u )} d u + \int sec (u) d u)

\int \frac{1}{sec ( u )} d u = sin (u)

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= \frac{1}{64} (- sin (u) + ln ∣ tan (u) + sec (u) ∣)

代用を戻す

u = arctan (\frac{4}{3} y)

= \frac{1}{64} (- sin (arctan (\frac{4}{3} y)) + ln tan (arctan (\frac{4}{3} y)) + sec (arctan (\frac{4}{3} y)))

簡素化

\frac{1}{64} (- sin (arctan (\frac{4}{3} y)) + ln tan (arctan (\frac{4}{3} y)) + sec (arctan (\frac{4}{3} y))) : \frac{1}{64} (- \frac{4 y}{( 9 + 16 y ^{2} ) ^{\frac{1}{2}}} + ln (\frac{1}{3} 4 y + (9 + 16 y^{2})^{\frac{1}{2}}))

= \frac{1}{64} (- \frac{4 y}{( 9 + 16 y ^{2} ) ^{\frac{1}{2}}} + ln (\frac{1}{3} 4 y + (9 + 16 y^{2})^{\frac{1}{2}}))

定数を解答に追加する

= \frac{1}{64} (- \frac{4 y}{( 9 + 16 y ^{2} ) ^{\frac{1}{2}}} + ln (\frac{1}{3} 4 y + (9 + 16 y^{2})^{\frac{1}{2}})) + C

グラフ

人気の例

integral of t 1/4

\int t \frac{1}{4} d t

integral of (6x+1)/(4x^2+4x+10)

\int \frac{6 x + 1}{4 x ^{2} + 4 x + 10} d x

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\int (2 x + 3 x^{2} + 1) d x

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\int [\frac{4 x ^{2} - 8 x + 4}{2 x - 2}] d x