integral of (ln(x))/(sqrt(1-x^2))

解答

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

i \frac{1}{2} (arcsin^{2} (x) + Li_{2} (e^{2 i a r c s i n (x)}) - arcsin (x) ln (1 - (- 1)^{\frac{2}{π} a r c s i n (x)}) + arcsin (x) ln (x)) + C

求解步骤

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

使用三角换元法

= \int ln (sin (u)) d u

这是非初等函数积分:

\int ln (sin (x)) d x = i \frac{1}{2} (x^{2} + Li_{2} (e^{2 i x}) - x ln (1 - (- 1)^{\frac{2 x}{π}}) + x ln (sin (x)))

= i \frac{1}{2} (u^{2} + Li_{2} (e^{2 i u}) - u ln (1 - (- 1)^{\frac{2 u}{π}}) + u ln (sin (u)))

u = arcsin (x)

代回

= i \frac{1}{2} (arcsin^{2} (x) + Li_{2} (e^{2 i a r c s i n (x)}) - arcsin (x) ln (1 - (- 1)^{\frac{2 a r c s i n ( x )}{π}}) + arcsin (x) ln (sin (arcsin (x))))

化简

i \frac{1}{2} (arcsin^{2} (x) + Li_{2} (e^{2 i a r c s i n (x)}) - arcsin (x) ln (1 - (- 1)^{\frac{2 a r c s i n ( x )}{π}}) + arcsin (x) ln (sin (arcsin (x)))) : i \frac{1}{2} (arcsin^{2} (x) + Li_{2} (e^{2 i a r c s i n (x)}) - arcsin (x) ln (1 - (- 1)^{\frac{2}{π} a r c s i n (x)}) + arcsin (x) ln (x))

= i \frac{1}{2} (arcsin^{2} (x) + Li_{2} (e^{2 i a r c s i n (x)}) - arcsin (x) ln (1 - (- 1)^{\frac{2}{π} a r c s i n (x)}) + arcsin (x) ln (x))

解答补常数

= i \frac{1}{2} (arcsin^{2} (x) + Li_{2} (e^{2 i a r c s i n (x)}) - arcsin (x) ln (1 - (- 1)^{\frac{2}{π} a r c s i n (x)}) + arcsin (x) ln (x)) + C

\int \frac{1}{27 - x ^{2}} d x

\int x^{\frac{- 4}{3}} d x

\int (3 + \frac{2}{5} x^{2} + \frac{3}{7} x^{3}) d x

\int \frac{x + 9}{64 - ( x - 8 ) ^{2}} d x

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