Zadatak 0044

Test: MPsIVKVIV_009

 

Reši integrale: a)  \(\int \frac{e^x}{\sqrt{9-e^{2x}}}dx\); b)  \(\int \frac{x^3-2x+2}{(x-1)^2(x^2+1)}dx\).

 

Rešenje:


 

a) \(\int \frac{e^x}{\sqrt{9-e^{2x}}}dx= \left ( e^x =t, e^xdx=dt\right )=\int \frac{dt}{\sqrt{9-t^2}}=\frac{1}{3}\int \frac{dt}{\sqrt{1-(\frac{t}{3})^2}}\)

 

\(=(\frac{t}{3}=u, dt=3du)=\int \frac{du}{\sqrt{1-u^2}}=\arcsin u=\arcsin \frac{t}{3}=\arcsin \frac{e^x}{3}+C\).

 

b) \(\frac{x^3-2x+2}{(x-1)^2(x^2+1)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+1}/\cdot (x-1)^2(x^2+1)\)

 

\(\Leftrightarrow x^3-2x+2=A(x-1)(x^2+1)+B(x^2+1)+(Cx+D)(x-1)^2\)

 

\(\Leftrightarrow x^3-2x+2=Ax^3+Ax-Ax^2-A+Bx^2+B+Cx^3-2Cx^2+Cx+Dx^2-2Dx+D\)

 

\(\Leftrightarrow x^3-2x+2=x^3(A+C)+x^2(-A+B-2C+D)+x(A+C-2D)+(-A+B+D)\)

 

Sada rešimo sistem jednačina:

 

\(A+C=1\)

\(-A+B-2C+D=0\)

\(A+C-2D=-2\)

\(-A+B+D=2\)

 

\(\Leftrightarrow A=0, B=2, C=1, D=\frac{3}{2}\)  \(\Leftrightarrow \frac{x^3-2x+2}{(x-1)^2(x^2+1)}=\frac{2}{(x-1)^2}+\frac{x+\frac{3}{2}}{x^2+1}\)

 

\(\int \frac{x^3-2x+2}{(x-1)^2(x^2+1)}dx\) \(=\int \frac{2}{(x-1)^2}dx+\frac{1}{2}\int \frac{2x+3}{1+x^2}dx\)

 

\(=-\frac{2}{x-1}+\frac{1}{2}\int \frac{2x}{1+x^2}dx+\frac{3}{2}\int \frac{1}{1+x^2}dx\)

 

\(=-\frac{2}{x-1}+\frac{1}{2}\ln (1+x^2)+\frac{3}{2}\arctan x+C\).