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人気のある微分積分 >

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

解

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

解

\frac{1}{2} (ln tan (\frac{arctan ( 2 x )}{2}) + 1 - ln tan (\frac{arctan ( 2 x )}{2}) - 1 - tan (\frac{arctan ( 2 x )}{2})) + C

解答ステップ

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

三角関数による置換を適用する

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

定数を除く:

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

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

簡素化

\frac{sec ^{2} ( u )}{1 + sec ( u )} : \frac{1}{cos ( u ) + cos ^{2} ( u )}

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

u置換積分法を適用する

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

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

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

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}{2} (\int \frac{2}{- v ^{2} + 1} d v - \int 1 d v)

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

\int 1 d v = v

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

代用を戻す

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

定数を解答に追加する

= \frac{1}{2} (ln tan (\frac{arctan ( 2 x )}{2}) + 1 - ln tan (\frac{arctan ( 2 x )}{2}) - 1 - tan (\frac{arctan ( 2 x )}{2})) + C

グラフ

人気の例

integral of x/(4sqrt(9-x^2))

\int \frac{x}{4 9 - x ^{2}} d x

integral of 1/(1+e^{11x)}

\int \frac{1}{1 + e ^{11 x}} d x

integral of (x+3)/(x^2+2x+10)

\int \frac{x + 3}{x ^{2} + 2 x + 10} d x

integral of 1/((1-2/9 x)^3)

\int \frac{1}{( 1 - \frac{2}{9} x ) ^{3}} d x

integral of rivative of sin(x)

\int \frac{d}{d x} (sin (x)) d x