zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的微积分 >

integral of 1/(sqrt(x^2+4/5))

解答

\int \frac{1}{x ^{2} + \frac{4}{5}} d x

解答

ln (\frac{1}{2} 5 x + 4 + 5 x^{2}) + C

求解步骤

\int \frac{1}{x ^{2} + \frac{4}{5}} d x

x^{2} + \frac{4}{5} = \frac{4}{5} (\frac{x ^{2}}{\frac{4}{5}} + 1)

= \int \frac{1}{\frac{4}{5} ( \frac{x ^{2}}{\frac{4}{5}} + 1 )} d x

\frac{1}{\frac{4}{5} ( \frac{x ^{2}}{\frac{4}{5}} + 1 )} = \frac{1}{\frac{4}{5}} \frac{1}{\frac{x ^{2}}{\frac{4}{5}} + 1}

= \int \frac{1}{\frac{4}{5}} \cdot \frac{1}{\frac{x ^{2}}{\frac{4}{5}} + 1} d x

提出常数:

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

= \frac{1}{\frac{4}{5}} \cdot \int \frac{1}{\frac{x ^{2}}{\frac{4}{5}} + 1} d x

\frac{1}{\frac{4}{5}} = \frac{5}{2}

= \frac{5}{2} \cdot \int \frac{1}{\frac{x ^{2}}{\frac{4}{5}} + 1} d x

使用三角换元法

= \frac{5}{2} \cdot \int \frac{2 sec ( u )}{5} d u

提出常数:

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

= \frac{5}{2} \cdot \frac{2}{5} \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{5}{2} \cdot \frac{2}{5} ln ∣ tan (u) + sec (u) ∣

u = arctan (\frac{5}{2} x)

代回

= \frac{5}{2} \cdot \frac{2}{5} ln tan (arctan (\frac{5}{2} x)) + sec (arctan (\frac{5}{2} x))

化简

\frac{5}{2} \cdot \frac{2}{5} ln tan (arctan (\frac{5}{2} x)) + sec (arctan (\frac{5}{2} x)) : ln (\frac{1}{2} 5 x + 4 + 5 x^{2})

= ln (\frac{1}{2} 5 x + 4 + 5 x^{2})

解答补常数

= ln (\frac{1}{2} 5 x + 4 + 5 x^{2}) + C

作图

流行的例子

integral of 1/((x+3)(x^2+x+1))

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

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

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

integral of (x^2)/((9-y^2)^{5/2)}

\int \frac{x ^{2}}{( 9 - y ^{2} ) ^{\frac{5}{2}}} d x

integral of (ln(x+2))/(2x+4)

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

integral of-y^2sin(x)

\int - y^{2} sin (x) d x