Question

Помогите с решением пожалуйста
интеграл ​

Answer 1

Ответ:

Замена в интегралах от иррациональных функций .

$\bf \displaystyle 13)\ \ \int \frac{dx}{\sqrt{x}\, (1+\sqrt[4]{\bf x^3})}=\Big[\ x=t^4\ ,\ dx=4t^3\, dt\ ,\ t=\sqrt[4]{x}\ \Big]=\\\\\\=\int \frac{4\, t^3\, dt}{t^2\, (1+t^3)}=4\int \frac{t\, dt}{1+t^3}=4\int \frac{t\, dt}{(1+t)(1-t+t^2)}=$  

Разложим дробь на сумму простейших дробей .

$\bf \displaystyle \frac{t}{(t+1)(t^2-t+1)}=\dfrac{A}{t+1}+\frac{Bt+C}{t^2-t+1}=\frac{At^2-At+A+Bt^2+Ct+Bt+C}{(t+1)(t^2-t+1)}\\\\\\t=(A+B)t^2+(B+C-A)t+(A+C)\\\\t^2\ |\ A+B=0\ \ ,\ \ \ \ \ \ \ \ \ A=-\frac{1}{3}\\t\ \ \, |\ B+C-A=1\ \ ,\ \ \ B=\frac{1}{3}\\t^0\ |\ A+C=0\ \ \ ,\qquad \ \ C=\frac{1}{3}$  

$\bf \displaystyle =-\frac{4}{3}\int \frac{dt}{t+1}+\frac{4}{3}\int \frac{(t+1)dt}{t^2-t+1}=-\frac{4}{3}\cdot ln\Big|t+1\, \Big|+\frac{4}{3}\int \frac{(t+1)\, dt}{(t-\frac{1}{2})^2+\frac{3}{4}}=\\\\\\=-\frac{4}{3}\cdot ln\Big|t+1\, \Big|+\frac{4}{3}\int \frac{(t-\frac{1}{2})+\frac{3}{2}}{(t-\frac{1}{2})^2+\frac{3}{4}}\, dt=$

$\bf \displaystyle =-\frac{4}{3}\cdot ln\Big|t+1\, \Big|+\frac{4}{3}\cdot \frac{1}{2}\int \frac{2(t-\frac{1}{2})\, dt}{(t-\frac{1}{2})^2+\frac{3}{4}}\, dt+\frac{4}{3}\cdot \frac{3}{2}\int \dfrac{dt}{(t-\frac{1}{2})^2+\frac{3}{4}}=\\\\\\=-\frac{4}{3}\cdot ln\Big|t+1\, \Big|+\frac{2}{3}\cdot ln\Big|\, (t-\frac{1}{2})^2+\frac{3}{4}\, \Big|+2\cdot \frac{2}{\sqrt3}\cdot arctg\frac{2\, (t-\frac{1}{2})}{\sqrt3}+C=$  

$\displaystyle \bf =-\frac{4}{3}\cdot ln\Big|\sqrt[4]{\bf x}+1\, \Big|+\frac{2}{3}\cdot ln\Big|\, (\sqrt[4]{\bf x}-\frac{1}{2})^2+\frac{3}{4}\, \Big|+2\cdot \frac{2}{\sqrt3}\cdot arctg\frac{2\, (\sqrt[4]{\bf x}-\frac{1}{2})}{\sqrt3}+C=\\\\\\=-\frac{4}{3}\cdot ln\Big|\sqrt[4]{\bf x}+1\, \Big|+\frac{2}{3}\cdot ln\Big|\, \sqrt{x}-\sqrt[4]{\bf x}+1\, \Big|+\frac{4}{\sqrt3}\cdot arctg\frac{2\sqrt[4]{\bf x}-1}{\sqrt3}+C$  

14)  Запишем подынтегральное выражение как квадратный корень из дроби .

$\displaystyle \bf \int \frac{\sqrt{a+x}}{\sqrt{a-x}}\, dx=\int \sqrt{\frac{a+x}{a-x}}\, dx=\Big[\ t^2=\frac{a+x}{a-x}\ \ \to \ \ at^2-xt^2=a+x\ ,\\\\\\at^2-a=xt^2+x\ \ ,\ \ a(t^2-1)=x(t^2+1)\ \ \to \ \ x=\frac{a(t^2-1)}{t^2+1}\ ,\\\\dx=a\cdot \frac{2t(t^2+1)-(t^2-1)\cdot 2t}{(t^2+1)^2}\, dt=a\cdot \frac{4\, t\, dt}{(t^2+1)^2}\ \Big]=\\\\\\=4a\int t\cdot \frac{t\, dt}{(t^2+1)^2}=4a\int \frac{t^2\, dt}{(t^2+1)^2}=4a\int \frac{t\cdot t\, dt}{(t^2+1)^2}=$  

Проинтегрируем по частям .

$\bf \displaystyle \bf =\Big[\ u=t\ ,\ du=dt\ ,\ dv=\frac{t\, dt}{(t^2+1)^2}\ ,\ v=-\frac{1}{2}\cdot \frac{1}{t^2+1}\ \Big]=\\\\\\=4a\left(-\frac{t}{2(t^2+1)}+\int \frac{dt}{2(t^2+1)}\right)=-\frac{4a\, t}{2(t^2+1)}+\frac{4a}{2}\cdot arctg\, t+C\right)=\\\\\\=-2a\cdot \sqrt{\frac{a+x}{a-x}}\cdot \frac{a+x}{2a}+2a\cdot arctg\sqrt{\frac{a+x}{a-x}}+C$