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人気のある微分積分 >

integral of sqrt(1+4x^2)

解

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

解

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

解答ステップ

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

三角関数による置換を適用する

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

積分換算を適用する

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

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

= \frac{1}{2} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

代用を戻す

u = arctan (2 x)

= \frac{1}{2} (\frac{sec ^{2} ( arctan ( 2 x ) ) sin ( arctan ( 2 x ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (2 x)) + sec (arctan (2 x)) ∣)

簡素化

\frac{1}{2} (\frac{sec ^{2} ( arctan ( 2 x ) ) sin ( arctan ( 2 x ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (2 x)) + sec (arctan (2 x)) ∣) : \frac{1}{4} (2 x 4 x^{2} + 1 + ln 2 x + 1 + 4 x^{2})

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

定数を解答に追加する

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

グラフ

人気の例

integral of x^2e^{x/2}

\int x^{2} e^{\frac{x}{2}} d x

integral of (3ln(x))/(x^2)

\int \frac{3 ln ( x )}{x ^{2}} d x

integral of (y^2)/y

\int \frac{y ^{2}}{y} d y

integral of sin^3(x)cos^{14}(x)

\int sin^{3} (x) cos^{14} (x) d x

limit as a approaches-1 of a

a \to - 1 lim (a)