integral of 1/(sqrt(16+4x^2))

解答

\int \frac{1}{16 + 4 x ^{2}} d x

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

求解步骤

\int \frac{1}{16 + 4 x ^{2}} d x

使用三角换元法

: \int \frac{sec ( u )}{2} d u

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

提出常数:

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

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

使用积分常见定则:

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

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

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

代回

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

化简

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

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

解答补常数

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

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

\int \frac{x}{9 + 25 x ^{4}} d x

\int \frac{4 x ^{2}}{( 8 x ^{3} - 4 )} d x

\int \frac{x ^{2} + x + 1}{x ^{3} + x ^{2}} d x

\int e^{0.1} d x