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人気のある微分積分 >

integral of xsqrt(1+1/(4x))

解

\int x 1 + \frac{1}{4 x} d x

解

\frac{1}{128} (8 (4 x - \frac{1}{2} + 1) x (4 x + 1) - ln 8 x + 1 + 4 x (4 x + 1)) + C

解答ステップ

\int x 1 + \frac{1}{4 x} d x

u置換積分法を適用する

= \int \frac{u u + 1}{16} d u

定数を除く:

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

= \frac{1}{16} \cdot \int u u + 1 d u

指数の規則を適用する:

a^{n} \cdot b^{n} = (a \cdot b)^{n}

u u + 1 = u (u + 1)

= \frac{1}{16} \cdot \int u (u + 1) d u

u置換積分法を適用する

= \frac{1}{16} \cdot \int (v - 1) v d v

拡張

(v - 1) v : v^{2} - v

= \frac{1}{16} \cdot \int v^{2} - v d v

平方完成する

v^{2} - v : (v - \frac{1}{2})^{2} - \frac{1}{4}

= \frac{1}{16} \cdot \int (v - \frac{1}{2})^{2} - \frac{1}{4} d v

u置換積分法を適用する

= \frac{1}{16} \cdot \int \frac{4 w ^{2} - 1}{2} d w

定数を除く:

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

= \frac{1}{16} \cdot \frac{1}{2} \cdot \int 4 w^{2} - 1 d w

三角関数による置換を適用する

= \frac{1}{16} \cdot \frac{1}{2} \cdot \int \frac{1}{2} tan^{2} (t) sec (t) d t

定数を除く:

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

= \frac{1}{16} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \int tan^{2} (t) sec (t) d t

三角関数の公式を使用して書き換える

= \frac{1}{16} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \int (- 1 + sec^{2} (t)) sec (t) d t

拡張

(- 1 + sec^{2} (t)) sec (t) : - sec (t) + sec^{3} (t)

= \frac{1}{16} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \int - sec (t) + sec^{3} (t) d t

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}{16} \cdot \frac{1}{2} \cdot \frac{1}{2} (- \int sec (t) d t + \int sec^{3} (t) d t)

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

\int sec^{3} (t) d t = \frac{1}{2} sec (t) tan (t) + \frac{1}{2} ln ∣ tan (t) + sec (t) ∣

= \frac{1}{16} \cdot \frac{1}{2} \cdot \frac{1}{2} (- ln ∣ tan (t) + sec (t) ∣ + \frac{1}{2} sec (t) tan (t) + \frac{1}{2} ln ∣ tan (t) + sec (t) ∣)

代用を戻す

= \frac{1}{16} \cdot \frac{1}{2} \cdot \frac{1}{2} (- ln tan (arcsec (2 (4 x + 1 - \frac{1}{2}))) + sec (arcsec (2 (4 x + 1 - \frac{1}{2}))) + \frac{1}{2} sec (arcsec (2 (4 x + 1 - \frac{1}{2}))) tan (arcsec (2 (4 x + 1 - \frac{1}{2}))) + \frac{1}{2} ln tan (arcsec (2 (4 x + 1 - \frac{1}{2}))) + sec (arcsec (2 (4 x + 1 - \frac{1}{2}))))

簡素化

\frac{1}{16} \cdot \frac{1}{2} \cdot \frac{1}{2} (- ln tan (arcsec (2 (4 x + 1 - \frac{1}{2}))) + sec (arcsec (2 (4 x + 1 - \frac{1}{2}))) + \frac{1}{2} sec (arcsec (2 (4 x + 1 - \frac{1}{2}))) tan (arcsec (2 (4 x + 1 - \frac{1}{2}))) + \frac{1}{2} ln tan (arcsec (2 (4 x + 1 - \frac{1}{2}))) + sec (arcsec (2 (4 x + 1 - \frac{1}{2})))) : \frac{1}{128} (8 (4 x - \frac{1}{2} + 1) x (4 x + 1) - ln 8 x + 1 + 4 x (4 x + 1))

= \frac{1}{128} (8 (4 x - \frac{1}{2} + 1) x (4 x + 1) - ln 8 x + 1 + 4 x (4 x + 1))

定数を解答に追加する

= \frac{1}{128} (8 (4 x - \frac{1}{2} + 1) x (4 x + 1) - ln 8 x + 1 + 4 x (4 x + 1)) + C

グラフ

人気の例

tangent sqrt(3x+25),\at x=-3

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\int (2 x - 1) ln (5 x) d x

(dP)/(dt)=Psqrt(t)

\frac{d P}{d t} = P t

integral of x^4ln(2x)

\int x^{4} ln (2 x) d x

derivative of x^{2ax}

\frac{d}{d x} (x^{2 a x})