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人気のある微分積分 >

integral of (sqrt(1+x^2))/(-9x)

解

\int \frac{1 + x ^{2}}{- 9 x} d x

解

\frac{1}{18} (ln 1 + x^{2} + 1 - ln 1 + x^{2} - 1 - 2 1 + x^{2}) + C

解答ステップ

\int \frac{1 + x ^{2}}{- 9 x} d x

定数を除く:

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

= \frac{1}{- 9} \cdot \int \frac{1 + x ^{2}}{x} d x

簡素化

= (- \frac{1}{9}) \cdot \int \frac{1 + x ^{2}}{x} d x

u置換積分法を適用する

= (- \frac{1}{9}) \cdot \int \frac{u ^{2}}{u ^{2} - 1} d u

因数

u^{2} - 1 : - (- u^{2} + 1)

= (- \frac{1}{9}) \cdot \int \frac{u ^{2}}{- ( - u ^{2} + 1 )} d u

定数を除く:

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

= (- \frac{1}{9}) (- \int \frac{u ^{2}}{- u ^{2} + 1} d u)

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

= (- \frac{1}{9}) (- \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

= (- \frac{1}{9}) (- (\int \frac{1}{- u ^{2} + 1} d u - \int 1 d u))

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

\int 1 d u = u

= (- \frac{1}{9}) (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2} - u))

代用を戻す

u = 1 + x^{2}

= (- \frac{1}{9}) (- (\frac{ln 1 + x ^{2} + 1}{2} - \frac{ln 1 + x ^{2} - 1}{2} - 1 + x^{2}))

簡素化

(- \frac{1}{9}) (- (\frac{ln 1 + x ^{2} + 1}{2} - \frac{ln 1 + x ^{2} - 1}{2} - 1 + x^{2})) : \frac{1}{18} (ln 1 + x^{2} + 1 - ln 1 + x^{2} - 1 - 2 1 + x^{2})

= \frac{1}{18} (ln 1 + x^{2} + 1 - ln 1 + x^{2} - 1 - 2 1 + x^{2})

定数を解答に追加する

= \frac{1}{18} (ln 1 + x^{2} + 1 - ln 1 + x^{2} - 1 - 2 1 + x^{2}) + C

グラフ

人気の例

integral of tan^2(y)

\int tan^{2} (y) d y

tangent y= 1/(x^3)

t an g e n t y = \frac{1}{x ^{3}}

integral from-2 to 2 of x^3-1

\int_{- 2}^{2} x^{3} - 1 d x

derivative z(v)=v^{3/2}(v+ce^v)

d er i v a t i v e z (v) = v^{\frac{3}{2}} (v + c e^{v})

y^{''}+y=(cot(x))^2

y^{^{''}} + y = (cot (x))^{2}