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人気のある微分積分 >

integral of sqrt(t^2+5)

解

\int t^{2} + 5 d t

解

5 (\frac{1}{10} t 5 + t^{2} + \frac{1}{2} ln \frac{1}{5} t + \frac{1}{5} (5 + t^{2})) + C

解答ステップ

\int t^{2} + 5 d t

三角関数による置換を適用する

= \int 5 sec^{3} (u) d u

定数を除く:

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

= 5 \cdot \int sec^{3} (u) d u

積分換算を適用する

= 5 (\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) ∣

= 5 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

代用を戻す

u = arctan (\frac{1}{5} t)

= 5 \frac{sec ^{2} ( arctan ( \frac{1}{5} t ) ) sin ( arctan ( \frac{1}{5} t ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{5} t)) + sec (arctan (\frac{1}{5} t))

簡素化

5 \frac{sec ^{2} ( arctan ( \frac{1}{5} t ) ) sin ( arctan ( \frac{1}{5} t ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{5} t)) + sec (arctan (\frac{1}{5} t)) : 5 (\frac{1}{10} t 5 + t^{2} + \frac{1}{2} ln \frac{1}{5} t + \frac{1}{5} (5 + t^{2}))

= 5 (\frac{1}{10} t 5 + t^{2} + \frac{1}{2} ln \frac{1}{5} t + \frac{1}{5} (5 + t^{2}))

定数を解答に追加する

= 5 (\frac{1}{10} t 5 + t^{2} + \frac{1}{2} ln \frac{1}{5} t + \frac{1}{5} (5 + t^{2})) + C

グラフ

人気の例

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

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

integral of 1/(cos(5x))

\int \frac{1}{cos ( 5 x )} d x

integral of 2/(4x)

\int \frac{2}{4 x} d x

integral of (2y+3)/((y-1)(y^3-1))

\int \frac{2 y + 3}{( y - 1 ) ( y ^{3} - 1 )} d y

integral of t(sec^2(t^2))(tan^4(t^2))

\int t (sec^{2} (t^{2})) (tan^{4} (t^{2})) d t