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人気のある微分積分 >

integral of (x^2)/(sqrt(x^2+4x-12))

解

\int \frac{x ^{2}}{x ^{2} + 4 x - 12} d x

解

\frac{1}{2} (x + 2) x^{2} + 4 x - 12 + 12 ln (\frac{1}{4} x + x^{2} + 4 x - 12 + 2) - 4 x^{2} + 4 x - 12 + C

解答ステップ

\int \frac{x ^{2}}{x ^{2} + 4 x - 12} d x

因数

\frac{x ^{2}}{x ^{2} + 4 x - 12}

= \int \frac{x ^{2}}{( x - 2 ) ( x + 6 )} d x

u置換積分法を適用する

= \int \frac{( u - 6 ) ^{2}}{u ( u - 8 )} d u

拡張

\frac{( u - 6 ) ^{2}}{u ( u - 8 )} : \frac{u ^{2}}{u ^{2} - 8 u} - \frac{12 u}{u ^{2} - 8 u} + \frac{36}{u ^{2} - 8 u}

= \int \frac{u ^{2}}{u ^{2} - 8 u} - \frac{12 u}{u ^{2} - 8 u} + \frac{36}{u ^{2} - 8 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

= \int \frac{u ^{2}}{u ^{2} - 8 u} d u - \int \frac{12 u}{u ^{2} - 8 u} d u + \int \frac{36}{u ^{2} - 8 u} d u

\int \frac{u ^{2}}{u ^{2} - 8 u} d u = \frac{1}{2} (u - 4) u^{2} - 8 u + 24 ln (\frac{1}{4} u + u^{2} - 8 u - 4) + 8 u^{2} - 8 u

\int \frac{12 u}{u ^{2} - 8 u} d u = 12 (u^{2} - 8 u + 4 ln (\frac{1}{4} u^{2} - 8 u + u - 4))

\int \frac{36}{u ^{2} - 8 u} d u = 36 ln (\frac{1}{4} u^{2} - 8 u + u - 4)

= \frac{1}{2} (u - 4) u^{2} - 8 u + 24 ln (\frac{1}{4} u + u^{2} - 8 u - 4) + 8 u^{2} - 8 u - 12 (u^{2} - 8 u + 4 ln (\frac{1}{4} u^{2} - 8 u + u - 4)) + 36 ln (\frac{1}{4} u^{2} - 8 u + u - 4)

代用を戻す

u = x + 6

= \frac{1}{2} (x + 6 - 4) (x + 6)^{2} - 8 (x + 6) + 24 ln (\frac{1}{4} x + 6 + (x + 6)^{2} - 8 (x + 6) - 4) + 8 (x + 6)^{2} - 8 (x + 6) - 12 ((x + 6)^{2} - 8 (x + 6) + 4 ln (\frac{1}{4} (x + 6)^{2} - 8 (x + 6) + x + 6 - 4)) + 36 ln (\frac{1}{4} (x + 6)^{2} - 8 (x + 6) + x + 6 - 4)

簡素化

\frac{1}{2} (x + 6 - 4) (x + 6)^{2} - 8 (x + 6) + 24 ln (\frac{1}{4} x + 6 + (x + 6)^{2} - 8 (x + 6) - 4) + 8 (x + 6)^{2} - 8 (x + 6) - 12 ((x + 6)^{2} - 8 (x + 6) + 4 ln (\frac{1}{4} (x + 6)^{2} - 8 (x + 6) + x + 6 - 4)) + 36 ln (\frac{1}{4} (x + 6)^{2} - 8 (x + 6) + x + 6 - 4) : \frac{1}{2} (x + 2) x^{2} + 4 x - 12 + 12 ln (\frac{1}{4} x + x^{2} + 4 x - 12 + 2) - 4 x^{2} + 4 x - 12

= \frac{1}{2} (x + 2) x^{2} + 4 x - 12 + 12 ln (\frac{1}{4} x + x^{2} + 4 x - 12 + 2) - 4 x^{2} + 4 x - 12

定数を解答に追加する

= \frac{1}{2} (x + 2) x^{2} + 4 x - 12 + 12 ln (\frac{1}{4} x + x^{2} + 4 x - 12 + 2) - 4 x^{2} + 4 x - 12 + C

グラフ

人気の例

integral of (8x^3+1/(2x^2))

\int (8 x^{3} + \frac{1}{2 x ^{2}}) d x

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in v erse l a pl a ce \frac{( 3 s - 4 )}{s ^{2} + 9}

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d/(dθ)(sec(θ)tan(θ))

\frac{d}{d θ} (sec (θ) tan (θ))

inverselaplace 3/(s-9)

in v erse l a pl a ce \frac{3}{s - 9}