integral of 1/(169-16x^2)

解答

\int \frac{1}{169 - 16 x ^{2}} d x

\frac{1}{104} (ln \frac{4}{13} x + 1 - ln \frac{4}{13} x - 1) + C

求解步骤

\int \frac{1}{169 - 16 x ^{2}} d x

使用换元积分法

= \int \frac{1}{52 ( - u ^{2} + 1 )} d u

提出常数:

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

= \frac{1}{52} \cdot \int \frac{1}{- u ^{2} + 1} d u

使用积分常见定则:

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= \frac{1}{52} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

u = \frac{4}{13} x

代回

= \frac{1}{52} (\frac{ln \frac{4}{13} x + 1}{2} - \frac{ln \frac{4}{13} x - 1}{2})

化简

\frac{1}{52} (\frac{ln \frac{4}{13} x + 1}{2} - \frac{ln \frac{4}{13} x - 1}{2}) : \frac{1}{104} (ln \frac{4}{13} x + 1 - ln \frac{4}{13} x - 1)

= \frac{1}{104} (ln \frac{4}{13} x + 1 - ln \frac{4}{13} x - 1)

解答补常数

= \frac{1}{104} (ln \frac{4}{13} x + 1 - ln \frac{4}{13} x - 1) + C

\frac{d}{d x} (e^{x a})

t y^{^{''}} + 4 y^{^{'}} = t^{2}

f f^{^{'}} (xx) = x x^{3} - x x^{2} - 6 xx

\frac{d}{d x} (e^{c o s h (x)} cos^{2} (x))

\int \frac{10 - 10 x}{1 - x} d x