User:Yorik sar

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[edit] new one

[edit] $\int \ln \left(x^2-1\right)\, dx$

$=x\ln\left(x^2-1\right) - \int \frac {x*2x}{x^2-1}\,dx = x\ln\left(x^2-1\right) - 2*\int \left( 1+\frac 1 {x^2-1} \right) \,dx = x\ln\left(x^2-1\right)-2x + 2\int\frac {dx} {x^2-1} =$

$= x\left(\ln\left(x^2-1\right)-2\right)+\int\frac{dx}{x-1}-\int\frac{dx}{x+1} =x\left(\ln\left(x^2-1\right)-2\right) + \ln\left|x-1\right| - \ln\left|x+1\right| = x\left(\ln\left(x^2-1\right)-2\right) + \ln\left|\frac{x-1}{x+1}\right|$

[edit] №3738

$I_1 = \int\limits_0^1 \cos\left(\ln\frac 1 x\right)\frac {x^b-x^a} {\ln x}\,dx = \int\limits_0^1dx \int\limits_a^bdy \cos\left(\ln\frac 1 x\right)x^y = \int\limits_a^bdy \int\limits_0^1dx \cos\left(\ln\frac 1 x\right)x^y$

$I = \int\limits_0^1 \cos\left(\ln\frac 1 x\right) x^y\,dx = \frac 1 {y+1}\int\limits_0^1 \cos\left(\ln\frac 1 x\right) d\left(x^{y+1}\right) = \frac 1 {y+1}\left( \left.\left( x^y\cos\ln\frac 1 x \right) \right|_0^1 - \int\limits_0^1 x^{y+1}\sin\ln\frac 1 x\,dx \right) =$ $= \frac 1 {y+1}\left( 1 - \frac 1 {y+1}\int\limits_0^1\sin\ln\frac 1 x d\left(x^{y+1}\right)\right) = \frac 1 {y+1} \left(1 - \frac 1 {y+1}\left(\left(\left.x^{y+1}\sin\ln\frac 1 x\right)\right|_0^1 + \int\limits_0^1\cos\left(\ln\frac 1 x\right) x^y\,dx\right)\right) = \frac 1 {y+1} - \frac 1 {(y+1)^2} I$ $I = \frac {y+1} {1+(y+1)^2}$