integral of sqrt(5x^2+1)

解答

\int 5 x^{2} + 1 d x

\frac{1}{2} x 1 + 5 x^{2} + \frac{1}{2 5} ln 5 x + 1 + 5 x^{2} + C

求解步骤

\int 5 x^{2} + 1 d x

使用三角换元法

= \int \frac{1}{5} sec^{3} (u) d u

提出常数:

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

= \frac{1}{5} \cdot \int sec^{3} (u) d u

使用降次积分法

= \frac{1}{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) ∣

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

u = arctan (5 x)

代回

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

化简

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

= \frac{1}{2} x 1 + 5 x^{2} + \frac{1}{2 5} ln 5 x + 1 + 5 x^{2}

解答补常数

= \frac{1}{2} x 1 + 5 x^{2} + \frac{1}{2 5} ln 5 x + 1 + 5 x^{2} + C

\int \frac{a e ^{x} + b}{a e ^{x} - b} d x

(9 x y^{2} - 4 y) + (9 x^{2} y - 4 x) y^{^{'}} = 0

\int (csc (3 x) - cot (3 x))^{2} d x

y^{^{''}} - 2 y^{^{'}} = sin (x), y (0) = 1, y^{^{'}} (0) = 2

\int t^{2} + 4 d t