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人気のある微分積分 >

integral of 1/((4x+1)sqrt(16x^2+4))

解

\int \frac{1}{( 4 x + 1 ) 16 x ^{2} + 4} d x

解

\frac{1}{4 5} ln \frac{tan ( \frac{a r c t a n ( 2 x )}{2} ) - 2}{5} + 1 - ln \frac{tan ( \frac{a r c t a n ( 2 x )}{2} ) - 2}{5} - 1 + C

解答ステップ

\int \frac{1}{( 4 x + 1 ) 16 x ^{2} + 4} d x

u置換積分法を適用する

= \int \frac{1}{4 u u ^{2} - 2 u + 5} d u

定数を除く:

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

= \frac{1}{4} \cdot \int \frac{1}{u u ^{2} - 2 u + 5} d u

平方完成する

u^{2} - 2 u + 5 : (u - 1)^{2} + 4

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

u置換積分法を適用する

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

三角関数による置換を適用する

= \frac{1}{4} \cdot \int \frac{sec ( w )}{2 tan ( w ) + 1} d w

u置換積分法を適用する

= \frac{1}{4} \cdot \int \frac{2}{4 t + 1 - t ^{2}} d t

定数を除く:

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

= \frac{1}{4} \cdot 2 \cdot \int \frac{1}{4 t + 1 - t ^{2}} d t

因数

4 t + 1 - t^{2} : - (- 4 t - 1 + t^{2})

= \frac{1}{4} \cdot 2 \cdot \int \frac{1}{- ( - 4 t - 1 + t ^{2} )} d t

定数を除く:

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

= \frac{1}{4} \cdot 2 (- \int \frac{1}{- 4 t - 1 + t ^{2}} d t)

平方完成する

- 4 t - 1 + t^{2} : (t - 2)^{2} - 5

= \frac{1}{4} \cdot 2 (- \int \frac{1}{( t - 2 ) ^{2} - 5} d t)

u置換積分法を適用する

= \frac{1}{4} \cdot 2 (- \int \frac{1}{r ^{2} - 5} d r)

置換積分法を適用する

= \frac{1}{4} \cdot 2 (- \int \frac{1}{5 ( s ^{2} - 1 )} d s)

定数を除く:

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

= \frac{1}{4} \cdot 2 (- \frac{1}{5} \cdot \int \frac{1}{s ^{2} - 1} d s)

因数

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

= \frac{1}{4} \cdot 2 (- \frac{1}{5} \cdot \int \frac{1}{- ( - s ^{2} + 1 )} d s)

定数を除く:

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

= \frac{1}{4} \cdot 2 (- \frac{1}{5} (- \int \frac{1}{- s ^{2} + 1} d s))

共通積分を使用する:

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

= \frac{1}{4} \cdot 2 (- \frac{1}{5} (- (\frac{ln ∣ s + 1 ∣}{2} - \frac{ln ∣ s - 1 ∣}{2})))

代用を戻す

= \frac{1}{4} \cdot 2 - \frac{1}{5} - \frac{ln \frac{t a n ( \frac{a r c t a n ( \frac{1}{2} ( 4 x + 1 - 1 ) )}{2} ) - 2}{5} + 1}{2} - \frac{ln \frac{t a n ( \frac{a r c t a n ( \frac{1}{2} ( 4 x + 1 - 1 ) )}{2} ) - 2}{5} - 1}{2}

簡素化

\frac{1}{4} \cdot 2 - \frac{1}{5} - \frac{ln \frac{t a n ( \frac{a r c t a n ( \frac{1}{2} ( 4 x + 1 - 1 ) )}{2} ) - 2}{5} + 1}{2} - \frac{ln \frac{t a n ( \frac{a r c t a n ( \frac{1}{2} ( 4 x + 1 - 1 ) )}{2} ) - 2}{5} - 1}{2} : \frac{1}{4 5} ln \frac{tan ( \frac{a r c t a n ( 2 x )}{2} ) - 2}{5} + 1 - ln \frac{tan ( \frac{a r c t a n ( 2 x )}{2} ) - 2}{5} - 1

= \frac{1}{4 5} ln \frac{tan ( \frac{a r c t a n ( 2 x )}{2} ) - 2}{5} + 1 - ln \frac{tan ( \frac{a r c t a n ( 2 x )}{2} ) - 2}{5} - 1

定数を解答に追加する

= \frac{1}{4 5} ln \frac{tan ( \frac{a r c t a n ( 2 x )}{2} ) - 2}{5} + 1 - ln \frac{tan ( \frac{a r c t a n ( 2 x )}{2} ) - 2}{5} - 1 + C

グラフ

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