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人気のある微分積分 >

integral of sqrt(16x^2+32)

解

\int 16 x^{2} + 32 d x

解

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

解答ステップ

\int 16 x^{2} + 32 d x

三角関数による置換を適用する

= \int 8 sec^{3} (u) d u

定数を除く:

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

= 8 \cdot \int sec^{3} (u) d u

積分換算を適用する

= 8 (\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) ∣

= 8 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

代用を戻す

u = arctan (\frac{4}{4 2} x)

= 8 \frac{sec ^{2} ( arctan ( \frac{4}{4 2} x ) ) sin ( arctan ( \frac{4}{4 2} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{4}{4 2} x)) + sec (arctan (\frac{4}{4 2} x))

簡素化

8 \frac{sec ^{2} ( arctan ( \frac{4}{4 2} x ) ) sin ( arctan ( \frac{4}{4 2} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{4}{4 2} x)) + sec (arctan (\frac{4}{4 2} x)) : 8 (\frac{1}{4} x 2 + x^{2} + \frac{1}{2} ln \frac{x}{2} + \frac{2 + x ^{2}}{2})

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

定数を解答に追加する

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

グラフ

人気の例

integral of 2/(2-x)

\int \frac{2}{2 - x} d x

integral of (-6)/(x(ln(x))^2)

\int \frac{- 6}{x ( ln ( x ) ) ^{2}} d x

integral of tan^3(t)sec(t)

\int tan^{3} (t) sec (t) d t

integral of 1/((x)^3)

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

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

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