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人気のある微分積分 >

integral of (x^3+16)/(x^2+16)

解

\int \frac{x ^{3} + 16}{x ^{2} + 16} d x

解

4 (2 (\frac{x ^{2}}{16} + 1 - ln \frac{x ^{2}}{16} + 1) + arctan (\frac{x}{4})) + C

解答ステップ

\int \frac{x ^{3} + 16}{x ^{2} + 16} d x

置換積分法を適用する

= \int \frac{4 ( 4 u ^{3} + 1 )}{u ^{2} + 1} d u

定数を除く:

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

= 4 \cdot \int \frac{4 u ^{3} + 1}{u ^{2} + 1} d u

拡張

\frac{4 u ^{3} + 1}{u ^{2} + 1} : \frac{4 u ^{3}}{u ^{2} + 1} + \frac{1}{u ^{2} + 1}

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

= 4 (\int \frac{4 u ^{3}}{u ^{2} + 1} d u + \int \frac{1}{u ^{2} + 1} d u)

\int \frac{4 u ^{3}}{u ^{2} + 1} d u = 2 (u^{2} + 1 - ln u^{2} + 1)

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

= 4 (2 (u^{2} + 1 - ln u^{2} + 1) + arctan (u))

代用を戻す

u = \frac{x}{4}

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

2 ((\frac{x}{4})^{2} + 1 - ln (\frac{x}{4})^{2} + 1) = 2 (\frac{x ^{2}}{16} + 1 - ln \frac{x ^{2}}{16} + 1)

= 4 (2 (\frac{x ^{2}}{16} + 1 - ln \frac{x ^{2}}{16} + 1) + arctan (\frac{x}{4}))

定数を解答に追加する

= 4 (2 (\frac{x ^{2}}{16} + 1 - ln \frac{x ^{2}}{16} + 1) + arctan (\frac{x}{4})) + C

グラフ

人気の例

x^2y^'+3xy=(sin(4x))/x

x^{2} y^{^{'}} + 3 x y = \frac{sin ( 4 x )}{x}

derivative of arcsin(xdx)

\frac{d}{d x} (arcsin (x) d x)

tangent f(x)=(x^3-25x)^8,(-5,0)

t an g e n t f (x) = (x^{3} - 25 x)^{8}, (- 5, 0)

integral of (2x)/(1-x)

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

integral of (-3x)/(sqrt((x-1)^2+4))

\int \frac{- 3 x}{( x - 1 ) ^{2} + 4} d x