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受欢迎的微积分 >

integral of 1/(sqrt(2x^2-7x+5))

解答

\int \frac{1}{2 x ^{2} - 7 x + 5} d x

解答

\frac{1}{2} ln (\frac{16 x ^{2} - 56 x + 40 + 4 x - 7}{3}) + C

求解步骤

\int \frac{1}{2 x ^{2} - 7 x + 5} d x

对

2 x^{2} - 7 x + 5

配方:

2 (x - \frac{7}{4})^{2} - \frac{9}{8}

= \int \frac{1}{2 ( x - \frac{7}{4} ) ^{2} - \frac{9}{8}} d x

使用换元积分法

= \int \frac{2 2}{16 u ^{2} - 9} d u

提出常数:

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

= 22 \cdot \int \frac{1}{16 u ^{2} - 9} d u

使用三角换元法

= 22 \cdot \int \frac{sec ( v )}{4} d v

提出常数:

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

= 22 \frac{1}{4} \cdot \int sec (v) d v

使用积分常见定则:

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

= 22 \frac{1}{4} ln ∣ tan (v) + sec (v) ∣

代回

= 22 \frac{1}{4} ln tan (arcsec (\frac{4}{3} (x - \frac{7}{4}))) + sec (arcsec (\frac{4}{3} (x - \frac{7}{4})))

化简

22 \frac{1}{4} ln tan (arcsec (\frac{4}{3} (x - \frac{7}{4}))) + sec (arcsec (\frac{4}{3} (x - \frac{7}{4}))) : \frac{1}{2} ln (\frac{16 x ^{2} - 56 x + 40 + 4 x - 7}{3})

= \frac{1}{2} ln (\frac{16 x ^{2} - 56 x + 40 + 4 x - 7}{3})

解答补常数

= \frac{1}{2} ln (\frac{16 x ^{2} - 56 x + 40 + 4 x - 7}{3}) + C

作图

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