Question

без решения, нужен только ответ!

Answer 1

Ответ:

Вычислить интеграл   $\bf \displaystyle \int \sqrt{\frac{1+x}{1-2x}}\cdot\frac{dx}{x}$   .

Замена  №3 :  $\bf \bf t=\sqrt{\dfrac{1+x}{1-2x}}\ \ \ \Rightarrow \ \ \ t^2=\dfrac{1+x}{1-2x}$   .

$\bf \displaystyle t^2(1-2x)=1+x\ \ ,\ \ \ t^2-2t^2x=1+x\ \ ,\ \ x+2t^2x=t^2-1\ \ ,\\\\x\, (1+2t^2)=t^2-1\ \ ,\ \ \ x=\frac{t^2-1}{1+2t^2}\ \ ,\\\\\\dx=\frac{2t(1+2t^2)-(t^2-1)\cdot 4t}{(1+2t^2)^2}\cdot dt=\frac{2t+4t^3-4t^3+4t}{(1+2t^2)^2}\cdot dt=\frac{6t\cdot dt}{(1+2t^2)^2}$  

$\bf \displaystyle \int \sqrt{\frac{1+x}{1-2x}}\cdot\frac{dx}{x}=\bf \displaystyle \int t\cdot \frac{6t}{(1+2t^2)^2} \cdot\frac{1+2t^2}{t^2-1}\cdot dt=\int \frac{6t^2\cdot dt}{(1+2t^2)(t-1)(t+1)}=$

Разложим дробь на сумму простейших дробей .

$\displaystyle \bf \frac{6t^2}{(2t^2+1)(t-1)(t+1)}=\frac{A}{t-1}+\frac{B}{t+1}+\frac{Ct+D}{2t^2+1}\ \ ;\\\\\\6t^2=A(t+1)(2t^2+1)+B(t-1)(2t^2+1)+(Ct+D)(t-1)(t+1)\ ;\\\\t=1:\ \ A=\frac{6t^2}{(2t^2+1)(t+1)}=\frac{6\cdot 1}{(2+1)(1+1)}=\frac{6}{6}=1\\\\\\t=-1:\ \ B=\frac{6t^2}{(2t^2+1)(t-1)}=\frac{6\cdot 1}{(2+1)(-1-1)}=\frac{6}{-6}=-1$  

$\bf 6t^2=A(t+1)(2t^2+1)+B(t-1)(2t^2+1)+(Ct+D)(t^2-1)\\\\6t^2=A(2t^3+t+2t^2+1)+B(2t^3+t-2t^2-1)+Ct^3-Ct+Dt^2-D\\\\6t^2=t^3(2A+2B+C)+t^2(2A-2B+D)+t(A+B-C)+(A-B-D)\\\\t^2\ |\ 6=2A-2B+D\ \ ,\ \ \ 6=2\cdot 1-2\cdot (-1)+D\ \ ,\ \ D=2\ ,\\{}\ t\, \ |\ 0=A+B-C\qquad ,\ \ 0=1+(-1)-C\qquad \ ,\ \ \quad \ C=0$  

$\bf \displaystyle \int \frac{6t^2\cdot dt}{(1+2t^2)(t-1)(t+1)}=\int \frac{dt}{t-1}-\int \frac{dt}{t+1}+\int \frac{2\, dt}{2t^2+1}=\\\\\\=ln|t-1|-ln|t+1|+\int \frac{dt}{t^2+\frac{1}{2}}=\\\\\\=ln|t-1|-ln|t+1|+\frac{1}{\frac{1}{\sqrt2}}\, arctg\frac{t}{\frac{1}{\sqrt2}}+C=\\\\\\=ln|t-1|-ln|t+1|+\sqrt2\, arctg(\sqrt2\, t)+C=\\\\\\=ln\,\Big|\, \sqrt{\frac{1+x}{1-2x}}-1\ \Big|-ln\,\Big|\, \sqrt{\frac{1+x}{1-2x}}+1\ \Big|+\sqrt2\, arctg\Big(\, \sqrt{\frac{2(1+x)}{1-2x}}\Big)+C$