integral of sqrt(4x^2+17)

解答

\int 4 x^{2} + 17 d x

\frac{1}{4} (2 x 4 x^{2} + 17 + 17 (ln 2 x + 17 \frac{4 x ^{2} + 17}{17} - \frac{ln ( 17 )}{2})) + C

求解步骤

\int 4 x^{2} + 17 d x

使用三角换元法

= \int \frac{17}{2} sec^{3} (u) d u

提出常数:

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

= \frac{17}{2} \cdot \int sec^{3} (u) d u

使用降次积分法

= \frac{17}{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{17}{2} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

u = arctan (\frac{2}{17} x)

代回

= \frac{17}{2} \frac{sec ^{2} ( arctan ( \frac{2}{17} x ) ) sin ( arctan ( \frac{2}{17} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{17} x)) + sec (arctan (\frac{2}{17} x))

化简

\frac{17}{2} \frac{sec ^{2} ( arctan ( \frac{2}{17} x ) ) sin ( arctan ( \frac{2}{17} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{17} x)) + sec (arctan (\frac{2}{17} x)) : \frac{1}{4} (2 x 4 x^{2} + 17 + 17 (ln 2 x + 17 \frac{4 x ^{2} + 17}{17} - \frac{ln ( 17 )}{2}))

= \frac{1}{4} (2 x 4 x^{2} + 17 + 17 (ln 2 x + 17 \frac{4 x ^{2} + 17}{17} - \frac{ln ( 17 )}{2}))

解答补常数

= \frac{1}{4} (2 x 4 x^{2} + 17 + 17 (ln 2 x + 17 \frac{4 x ^{2} + 17}{17} - \frac{ln ( 17 )}{2})) + C

\int t sinh (t^{2}) d t

\int \frac{1}{1 - sec ( t )} d t

\int \frac{e ^{x}}{e ^{2 x} + 12 e ^{x} + 27} d x

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

\int (6 z^{2}) d z